Probability — Important Questions

• 40 students own dogs.
• 30 students own cats.
• 20 students own birds.
• 10 students own fish.
• 5 students own other pets.
• 15 students own both dogs and cats.
• 8 students own both cats and birds.
• 5 students own both dogs and birds.
• 2 students own dogs, cats and birds.
  Assuming that each student owns at least one pet, answer the following questions:
  (i) What is the probability that a randomly selected student owns a dog? [1 mark]
  (ii) What is the probability that a randomly selected student owns a cat or a bird? [2 marks]
  (iii) What is the probability that a randomly selected student owns a dog, a cat, and a bird? [1 mark]
  (iv) What is the probability that a randomly selected student owns exactly one type of pet? [1 mark]
  20. A lottery contains 10 prizes and 25 blanks. A ticket is drawn at random. What is the probability of winning a prize? [2 marks]

Answers

This implies $x=3$.
Number of blue balls = 3.

If the question implies a standard 52-card deck and removing 2 joker cards:
Total cards = 52. Face cards = 12. Probability = 12/52 = 3/13.

Let's interpret "A joker card is removed from a deck of 52 cards" as a standard 52-card deck, and one additional joker card is considered (which is not standard).
If a joker card is removed, the remaining cards are 51.
However, joker cards are not typically part of a standard 52-card deck for probability questions unless specified. Assuming it's a standard deck and the "joker card" is a mistake or refers to something else, like removing a specific card.

Re-interpreting: "A joker card is removed from a deck of 52 cards" is ambiguous. Let's assume it means one card (say, a specific one like Jack of Spades) is removed from the 52 cards.
Remaining cards = 51.
Face cards (J, Q, K) in a standard deck = 12.
If the removed card was a face card, the number of face cards becomes 11. P = 11/51.
If the removed card was not a face card, the number of face cards remains 12. P = 12/51.

Let's assume the question meant to say: "From a deck of 52 cards, one card is removed at random. Then another card is drawn. What is the probability of drawing a face card?". This is not what is written.

Let's assume the question implies removing two joker cards from a deck that originally had 52 cards + 2 jokers. So, 52 cards remain.
If the question truly implies a standard 52-card deck and one card is removed:
If the removed card is a face card, then there are 11 face cards left out of 51. P = 11/51.
If the removed card is NOT a face card, then there are 12 face cards left out of 51. P = 12/51.

Let's assume a simpler interpretation often seen in school contexts: "A deck of 52 cards has some cards removed. What is the probability...". If a joker card is removed, it implies the deck might have had jokers. Let's assume it's a standard 52-card deck and the statement is flawed. If we ignore the "joker card is removed" part and just consider "a deck of 52 cards", then P(face card) = 12/52 = 3/13.

Given the context of other questions, it's likely expecting a standard deck. The "joker card removed" might be a typo or a poorly phrased question. If we assume a standard 52 card deck and one specific card (e.g., Jack of Spades) is removed, and we want the probability of drawing any face card from the remaining 51.

Let's consider the possibility that the question means: "From a deck of 52 cards, the two Joker cards are removed, leaving 52 cards. What is the probability of drawing a face card?" This doesn't make sense as removing jokers increases the deck size.

Let's assume the question meant: "A deck of 52 cards. What is the probability of drawing a face card IF the Jack of Spades is removed?".
Total cards = 51.
Face cards = 11 (J, Q, K of 4 suits minus J of Spades).
Probability = 11/51.

Let's assume the question meant: "From a deck of 52 cards, one card is removed at random. What is the probability that the removed card was a face card?". This would be 12/52 = 3/13.

Let's try a very common phrasing: "A card is drawn from a standard deck of 52 cards. What is the probability of it being a face card?" Answer: 12/52 = 3/13.

The prompt asks to generate questions similar to past papers. Past papers often deal with standard decks. Let's assume the question means a standard 52-card deck, and the removal of a "joker card" is a misunderstanding or a poorly worded detail. If we have to make sense of it, perhaps it means 52 cards in total, and one card that ISN'T a face card is removed.

Let's assume the intent was: "A standard deck of 52 cards. Remove one specific card which is NOT a face card. What is the probability of drawing a face card from the remaining cards?"
Total cards = 51. Number of face cards = 12. Probability = 12/51 = 4/17.

Let's consider the simplest interpretation: The question is about a standard 52-card deck. The statement "A joker card is removed from a deck of 52 cards" is confusing. If it implies that there were originally 53 cards (52 + 1 joker) and the joker is removed, we are left with 52 cards. Then the probability of drawing a face card is 12/52 = 3/13. This seems the most plausible interpretation for a 2-mark question.

Let's assume the question meant: "From a deck of 52 cards, one card is removed. What is the probability that the remaining 51 cards contain 12 face cards?" This is also not the question.

Given the past paper style, the question likely intends to ask about drawing a face card from a standard deck. Let's assume the "joker card removed" implies a modification.
If we consider the removal of jokers from a deck that might have had them (e.g., 54 cards total). If 2 jokers are removed, we have 52 cards.
Then P(Face Card) = 12/52 = 3/13.

Let's go with the most straightforward interpretation of the question as it is written, even if it's slightly ambiguous.
A deck of 52 cards. One card (a joker) is removed. This implies the total cards become 51.
If this joker was a face card, there are now 11 face cards. P = 11/51.
If this joker was NOT a face card, there are still 12 face cards. P = 12/51.
This is too complex for a 2-mark question.

Let's assume it means: "Consider a standard deck of 52 cards. If one card is removed such that the number of face cards is affected, what is the probability of drawing a face card from the rest?". This is also too complex.

Let's assume the question meant: "From a standard deck of 52 cards, all the Aces and Kings are removed (these are 8 cards). What is the probability of drawing a face card from the remaining cards?"
Remaining cards = 52 - 8 = 44.
Face cards are J, Q, K. Kings are removed. So J and Q are left. There are 4 Jacks and 4 Queens.
Total face cards remaining = 4 + 4 = 8.
Probability = 8/44 = 2/11. This aligns with past paper question 14.

However, question 14 is 2 marks. This question is also 2 marks. The phrasing is slightly different.
Let's go back to the original phrasing: "A joker card is removed from a deck of 52 cards." This implies the original deck has 52 cards. A joker is NOT a standard card in a 52-card deck. This means the question is likely flawed.
If we assume the question means: "A deck of 52 cards is used, and one specific card (call it a 'joker' conceptually) is removed. What is the probability of drawing a face card from the remaining 51 cards?"
Let's assume the removed card was NOT a face card. Then there are 12 face cards out of 51. Probability = 12/51 = 4/17.

Let's re-read past paper 13: "The king, queen and ace of clubs and diamonds are removed from a deck of 52 playing cards...". That's 3*2 = 6 cards removed. Remaining = 46.
Past paper 14: "From a pack of 52 cards, all aces and all kings are removed...". That's 4+4 = 8 cards removed. Remaining = 44.

The question states "A joker card is removed from a deck of 52 cards".
Possibility 1: The deck has 52 cards, one of which is a joker. This is non-standard. If the joker is not a face card, P(face card) = 12/51. If joker is a face card (unlikely), P(face card) = 11/51.
Possibility 2: The deck has 52 standard cards. A joker card (which is separate) is also considered, making it 53 cards. Then the joker is removed, leaving 52 cards. P(face card) = 12/52 = 3/13.
Possibility 3: The question is poorly phrased and refers to a standard 52-card deck, and the 'joker' removal is a distractor or error. P(face card) = 12/52 = 3/13.

Given the mark allocation (2 marks), the simplest interpretation is usually intended. Let's assume Possibility 2 or 3, leading to 3/13. However, the phrasing "removed from a deck of 52 cards" suggests the deck itself is reduced in size.

Let's assume it implies a standard 52-card deck, and a card that is NOT a face card is removed. So, 51 cards remain, 12 of which are face cards. Probability = 12/51 = 4/17. This is plausible.

Let's try another interpretation for 2 marks: what if "joker card" is meant to represent one of the face cards? For example, the Jack of Spades.
If Jack of Spades is removed, 51 cards remain. Number of face cards = 11. P = 11/51.

Let's consider past paper 13 and 14 again. They remove multiple cards. This question removes ONE card.
If a non-face card is removed: 51 cards, 12 face cards. P = 12/51 = 4/17.
If a face card is removed: 51 cards, 11 face cards. P = 11/51.

Since the question does not specify which card is removed, and it's a simple removal, the simplest case is a non-face card.
Final decision: Assume a non-face card is removed from a standard 52-card deck.
Total cards = 52. Number of face cards (J, Q, K) = 12. Number of non-face cards = 40.
One non-face card is removed.
Remaining cards = 51.
Number of face cards remaining = 12.
Probability = 12/51 = 4/17.

Let's re-read: "A jar containing only coloured balls is 3/5". This implies all balls in the jar are coloured, and the probability of picking one is 3/5. This is contradictory.
Let's assume the question meant: "A jar contains some balls. The probability of selecting a coloured ball is 3/5. There are 20 coloured balls. Find the number of non-coloured balls."

If P(Coloured) = 3/5, and Number of Coloured balls = 20.
Let Total balls = T.
P(Coloured) = Number of Coloured balls / Total balls
3/5 = 20 / T
3T = 100
T = 100/3. Still not an integer.

Let's assume the question meant: "A jar contains 20 balls, some of which are coloured and some are not. The probability of selecting a coloured ball is 3/5."
Total balls = 20.
P(Coloured) = 3/5.
Number of coloured balls = P(Coloured) * Total balls = (3/5) * 20 = 12.
Number of non-coloured balls = Total balls - Number of coloured balls = 20 - 12 = 8.
This makes more sense and is a typical question format. However, the original wording states "A jar containing only coloured balls...".

Let's assume the wording of the original question is precisely as intended and try to resolve the contradiction.
"A jar contains some balls. The probability of selecting a coloured ball at random from the jar is 3/5. If there are 20 coloured balls in the jar, find the number of balls that are not coloured."
The statement "containing only coloured balls" is the source of confusion. If we ignore it, and treat "coloured balls" as a specific category within the jar:
Let $N_{col}$ be the number of coloured balls, and $N_{total}$ be the total number of balls.
$P(\text{Coloured})=\frac{N_{col}}{N_{total}}=\frac{3}{5}$.
We are given $N_{col}=20$.
So, $\frac{20}{N_{total}}=\frac{3}{5}$.
$3\times N_{total}=20\times 5=100$.
$N_{total}=\frac{100}{3}$. This is still not an integer.

Let's assume there's a typo in the numbers. If P(Coloured) = 2/3 and N_col = 20, then $N_{total}=30$. Non-coloured = 10.
If P(Coloured) = 4/5 and N_col = 20, then $N_{total}=25$. Non-coloured = 5.

Let's try to rephrase it to fit the 3-mark requirement and create a solvable problem.
"A jar contains red and blue balls. The probability of selecting a red ball at random is 2/5. If the jar contains 12 blue balls, find the total number of balls in the jar and the probability of selecting a red ball."
P(Red) = 2/5. P(Blue) = 1 - 2/5 = 3/5.
Number of blue balls = 12.
P(Blue) = Number of blue balls / Total balls
3/5 = 12 / Total balls
Total balls = (12 * 5) / 3 = 20.
Number of red balls = 20 - 12 = 8.
Probability of selecting a red ball = 8/20 = 2/5. This is consistent.

Let's stick to the original question's structure and numbers as much as possible, assuming a typo in the question itself. The most common type of problem is where the number of balls of a certain type is given, and the total number of balls needs to be found, or vice versa. The contradiction suggests the problem as stated is unsolvable with integer numbers of balls.

However, if we must answer it, we have to acknowledge the non-integer result for total balls.
If $N_{total}=\frac{100}{3}$, then the number of non-coloured balls = $N_{total}-N_{col}=\frac{100}{3}-20=\frac{100-60}{3}=\frac{40}{3}$.
This is highly unlikely for an exam question.

Let's assume the question meant: "A jar contains some balls, some of which are blue, and the rest are green. The probability of selecting a blue ball is 3/5. If there are 20 blue balls, how many green balls are there?"
P(Blue) = 3/5.
Number of blue balls = 20.
Let Total balls = T.
P(Blue) = 20/T = 3/5 => 3T = 100 => T = 100/3. Still the same issue.

Let's try reversing the ratio: "The probability of selecting a blue ball is 5/3" - impossible.
What if the number of coloured balls is not 20, but related to the fraction?
"A jar contains some balls. The probability of selecting a coloured ball at random is 3/5. If there are 20 balls which are not coloured, find the number of balls that are coloured."
P(Coloured) = 3/5.
P(Not Coloured) = 1 - 3/5 = 2/5.
Number of not coloured balls = 20.
P(Not Coloured) = Number of not coloured balls / Total balls
2/5 = 20 / Total balls
Total balls = (20 * 5) / 2 = 50.
Number of coloured balls = Total balls - Number of not coloured balls = 50 - 20 = 30.
This is a solvable problem and fits the 3-mark style. Let's use this interpretation.

Revised interpretation for Q9: A jar contains some balls, some coloured and some not. The probability of selecting a coloured ball at random is 3/5. If there are 20 balls that are NOT coloured, find the number of coloured balls in the jar.
P(Coloured) = 3/5.
P(Not Coloured) = 1 - P(Coloured) = 1 - 3/5 = 2/5.
Number of not coloured balls = 20.
Let Total number of balls = T.
P(Not Coloured) = $\frac{\text{Number of not coloured balls}}{\text{Total number of balls}}$

$T=50$.
Number of coloured balls = Total balls - Number of not coloured balls = 50 - 20 = 30.
The number of coloured balls is 30.

(ii) If 6 more blue balls are added, the number of blue balls becomes $m+6$.
The total number of balls becomes $30+6=36$.
The probability of drawing a blue ball in this case is $P(\text{Blue, new})=\frac{m+6}{36}$.
The original probability of drawing a blue ball was $P(\text{Blue, old})=\frac{m}{30}$.
According to the question, $P(\text{Blue, new})=\frac{5}{4}\times P(\text{Blue, old})$.
So, $\frac{m+6}{36}=\frac{5}{4}\times \frac{m}{30}$.
$\frac{m+6}{36}=\frac{5m}{120}$.
Simplify $\frac{5}{120}$ to $\frac{1}{24}$.
$\frac{m+6}{36}=\frac{m}{24}$.
Cross-multiply: $24(m+6)=36m$.
$24m+144=36m$.
$144=36m-24m$.
$144=12m$.
$m=\frac{144}{12}=12$.
So, the value of $m$ is 12.

(i) Cards of clubs remaining: 2, 3, 4, 5, 6, 7, 8, 9, 10, J. (10 cards).
Original clubs had 13 cards. K♣, Q♣, A♣ were removed. So 10 cards remain.
Number of favourable outcomes (card of clubs) = 10.
Total possible outcomes = 46.
Probability (a card of clubs) = $\frac{10}{46}=\frac{5}{23}$.

(ii) Red coloured cards: Hearts and Diamonds.
Original Hearts: 13 cards (all remain).
Original Diamonds: 13 cards. K♦, Q♦, A♦ were removed. So $13-3=10$ diamonds remain.
Total red coloured cards remaining = 13 (Hearts) + 10 (Diamonds) = 23.
Total possible outcomes = 46.
Probability (a red coloured card) = $\frac{23}{46}=\frac{1}{2}$.

(i) Face cards are Jack, Queen, King. Kings are removed. So, face cards remaining are Jacks and Queens.
Number of Jacks = 4.
Number of Queens = 4.
Total number of face cards remaining = $4+4=8$.
Probability (a face card) = $\frac{8}{44}=\frac{2}{11}$.

(ii) Card of red colour: Hearts and Diamonds.
Original Hearts: 13 cards. (No Aces or Kings removed from Hearts in this question). So 13 Hearts remain.
Original Diamonds: 13 cards. Ace of Diamonds and King of Diamonds were removed. So $13-2=11$ Diamonds remain.
Total red cards remaining = 13 (Hearts) + 11 (Diamonds) = 24.
Probability (a card of red colour) = $\frac{24}{44}=\frac{6}{11}$.

(ii) Making Purple: Blue and red make purple. This means one spinner shows Red and the other shows Blue.
The outcomes for 'Making Purple' are (R, B) and (B, R) - wait, Spinner II has Blue, but Spinner I does not.
Let's list them again clearly:
Spinner I: R, G, Y
Spinner II: R, B, G

Outcomes:
(R, R), (R, B), (R, G)
(G, R), (G, B), (G, G)
(Y, R), (Y, B), (Y, G)

'Making Purple' requires one spinner showing Red and the other showing Blue.
Possible outcomes: (R from Spinner I, B from Spinner II) => (R, B)
And (B from Spinner II, R from Spinner I) => This requires Spinner I to have Blue. It doesn't.
The condition is "Blue and red make purple. So, if one spinner shows Red (R) and another Blue(B), then you 'win'."
This means the pair must contain an R and a B.
The pairs that satisfy this are: (R, B) and (Y, B). Wait, (R,B) is the only one with R and B.
No, this is also incorrect.
The winning condition is: Spinner I shows Red AND Spinner II shows Blue OR Spinner I shows Blue AND Spinner II shows Red.
Spinner I outcomes: R, G, Y
Spinner II outcomes: R, B, G
Let's check pairs for R and B:

• (R from I, B from II) => (R, B) - this is a win.
• (B from I, R from II) => Spinner I does not have B. So this combination is not possible.
• What if Spinner I shows Red, and Spinner II shows... ?
  Let's assume the question meant that if one spinner is Red and another is Blue, regardless of which spinner shows which colour.
  The pairs are (R, B) and (Y, B). No, this is not correct.
  The outcomes are: (R,R), (R,B), (R,G), (G,R), (G,B), (G,G), (Y,R), (Y,B), (Y,G).
  Which of these contains one Red and one Blue?
• (R, B) : Spinner I is Red, Spinner II is Blue. This is a win.
• Is there any other pair? No. The outcomes are unique pairs.

Let's re-read: "Blue and red make purple. So, if one spinner shows Red (R) and another Blue(B), then you 'win'."
This implies we need to find pairs (X, Y) where X is an outcome from Spinner I, and Y is an outcome from Spinner II, such that {X, Y} = {Red, Blue}.
Possible pairs:

• Spinner I shows Red, Spinner II shows Blue: (R, B)
• Spinner I shows Blue, Spinner II shows Red: This is not possible because Spinner I does not have Blue.

So, the only outcome that makes purple is (R, B).
Probability of 'Making Purple' = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}=\frac{1}{9}$.

(iii) (a) For each win, a participant gets ₹ 10. If they lose, they pay ₹ 5.
Number of participants = 99.
Number of winning outcomes = 1 (Making Purple).
Number of losing outcomes = Total outcomes - Winning outcomes = 9 - 1 = 8.
Probability of winning = 1/9. Probability of losing = 8/9.
Expected winnings for 99 participants:
Number of wins = $99\times P(\text{Win})=99\times \frac{1}{9}=11$ wins.
Number of losses = $99\times P(\text{Lose})=99\times \frac{8}{9}=88$ losses.
Total money received by participants = $11\times \text{₹}10=\text{₹}110$.
Total money paid by participants = $88\times \text{₹}5=\text{₹}440$.
Total fund collected by school = Total money paid by participants - Total money received by participants
= $\text{₹}440-\text{₹}110=\text{₹}330$.

(iii) (b) If the same amount of ₹ 5 has been decided for winning or losing.
Number of wins = 11. Money received by participants = $11\times \text{₹}5=\text{₹}55$.
Number of losses = 88. Money paid by participants = $88\times \text{₹}5=\text{₹}440$.
Total fund collected by school = Total money paid - Total money received
= $\text{₹}440-\text{₹}55=\text{₹}385$.

(i) Probability (drawn ball is red) = $\frac{\text{Number of red balls}}{\text{Total number of balls}}=\frac{4}{9}$.

(ii) Probability (drawn ball is yellow) = $\frac{\text{Number of yellow balls}}{\text{Total number of balls}}=\frac{2}{9}$.

The event is "neither red nor black". This means the ball drawn must be white.
Number of white balls = 5.
Probability (ball is neither red nor black) = Probability (ball is white) = $\frac{\text{Number of white balls}}{\text{Total number of balls}}=\frac{5}{15}=\frac{1}{3}$.

Q15 - Detailed Answer for Part (iii) (a) and (b)

The game involves two spinners.
Spinner I outcomes: {R, G, Y} (3 outcomes)
Spinner II outcomes: {R, B, G} (3 outcomes)

Total possible outcomes when spinning both spinners = $3\times 3=9$.
These are: (R,R), (R,B), (R,G), (G,R), (G,B), (G,G), (Y,R), (Y,B), (Y,G).
All these outcomes are equally likely.

Winning condition: 'Making Purple' if one spinner shows Red (R) and another shows Blue (B).
The only outcome satisfying this is (R, B). So, there is 1 winning outcome.
The probability of winning P(Win) = $\frac{1}{9}$.

The probability of losing P(Lose) = $1-P(\text{Win})=1-\frac{1}{9}=\frac{8}{9}$.

There are 99 participants.

(iii) (a) Fund collection if win = ₹ 10, lose = ₹ 5
Number of participants who win = $99\times P(\text{Win})=99\times \frac{1}{9}=11$.
Total amount received by winners = $11\times \text{₹}10=\text{₹}110$.

Number of participants who lose = $99\times P(\text{Lose})=99\times \frac{8}{9}=88$.
Total amount paid by losers = $88\times \text{₹}5=\text{₹}440$.

The school collects money from losers and pays money to winners.
Total fund collected by the school = Total money paid by losers - Total money received by winners
= $\text{₹}440-\text{₹}110=\text{₹}330$.

(iii) (b) Fund collection if win = ₹ 5, lose = ₹ 5
Number of participants who win = 11.
Total amount received by winners = $11\times \text{₹}5=\text{₹}55$.

Number of participants who lose = 88.
Total amount paid by losers = $88\times \text{₹}5=\text{₹}440$.

Total fund collected by the school = Total money paid by losers - Total money received by winners
= $\text{₹}440-\text{₹}55=\text{₹}385$.

Q19 - Detailed Answer

The question asks for probabilities related to a survey of 100 students and their pets.
Total number of students surveyed = 100.

Data given:

• Dogs: 40
• Cats: 30
• Birds: 20
• Fish: 10
• Other pets: 5
• Dogs AND Cats: 15
• Cats AND Birds: 8
• Dogs AND Birds: 5
• Dogs AND Cats AND Birds: 2

It is stated that "each student owns at least one pet". This means we are dealing with sets and their overlaps.

(i) Probability that a randomly selected student owns a dog?
Number of students who own dogs = 40.
Total number of students = 100.
P(Dog) = $\frac{\text{Number of students who own dogs}}{\text{Total number of students}}=\frac{40}{100}=\frac{2}{5}$.

(ii) Probability that a randomly selected student owns a cat or a bird?
This requires the use of the Principle of Inclusion-Exclusion for two sets:
P(Cat or Bird) = P(Cat) + P(Bird) - P(Cat and Bird)
P(Cat) = $\frac{30}{100}$
P(Bird) = $\frac{20}{100}$
P(Cat and Bird) = $\frac{\text{Number of students who own cats AND birds}}{\text{Total number of students}}=\frac{8}{100}$

P(Cat or Bird) = $\frac{30}{100}+\frac{20}{100}-\frac{8}{100}=\frac{30+20-8}{100}=\frac{42}{100}=\frac{21}{50}$.

(iii) Probability that a randomly selected student owns a dog, a cat, and a bird?
This is directly given in the data.
Number of students who own dogs AND cats AND birds = 2.
Total number of students = 100.
P(Dog and Cat and Bird) = $\frac{2}{100}=\frac{1}{50}$.

(iv) Probability that a randomly selected student owns exactly one type of pet?
To find this, we need to calculate the number of students who own only one type of pet. We can use a Venn diagram approach or formulas.

Let D = set of students owning dogs, C = set of students owning cats, B = set of students owning birds.
We are given:
|D| = 40
|C| = 30
|B| = 20
|D ∩ C| = 15
|C ∩ B| = 8
|D ∩ B| = 5
|D ∩ C ∩ B| = 2

Number of students owning Dogs ONLY:
|D only| = |D| - |D ∩ C| - |D ∩ B| + |D ∩ C ∩ B|
This formula is for |D ∩ C' ∩ B'|.
A better way:
|D only| = |D| - (|D ∩ C| - |D ∩ C ∩ B|) - (|D ∩ B| - |D ∩ C ∩ B|) - |D ∩ C ∩ B|
|D only| = |D| - |(D ∩ C) \ (D ∩ C ∩ B)| - |(D ∩ B) \ (D ∩ C ∩ B)| - |D ∩ C ∩ B|
|D only| = |D| - |D ∩ C only| - |D ∩ B only| - |D ∩ C ∩ B|

Let's calculate the specific intersections first:

• Only D, C, B: |D ∩ C ∩ B| = 2
• Only D and C (not B): |D ∩ C| - |D ∩ C ∩ B| = 15 - 2 = 13
• Only C and B (not D): |C ∩ B| - |D ∩ C ∩ B| = 8 - 2 = 6
• Only D and B (not C): |D ∩ B| - |D ∩ C ∩ B| = 5 - 2 = 3

Now, calculate those owning only one pet:

• Only Dogs: |D only| = |D| - (Only D&C) - (Only D&B) - (D&C&B)
  |D only| = 40 - 13 - 3 - 2 = 40 - 18 = 22.
• Only Cats: |C only| = |C| - (Only D&C) - (Only C&B) - (D&C&B)
  |C only| = 30 - 13 - 6 - 2 = 30 - 21 = 9.
• Only Birds: |B only| = |B| - (Only C&B) - (Only D&B) - (D&C&B)
  |B only| = 20 - 6 - 3 - 2 = 20 - 11 = 9.

These calculations are for pets D, C, B. We also have "Fish" and "Other pets".
The problem states "each student owns at least one pet". This implies that the total number of students in the union of ALL pet ownerships is 100.
However, the data provided only covers D, C, B overlaps. It is not clear how Fish and Other pets fit in.

Let's re-read the question part (iv): "What is the probability that a randomly selected student owns exactly one type of pet?"
If "type of pet" refers to the categories mentioned (Dog, Cat, Bird, Fish, Other), then we need the counts for "Only Fish" and "Only Other".
The question structure implies D, C, B are the primary sets for calculation, and Fish/Other are distinct single categories.

If "exactly one type of pet" means strictly only ONE type among ALL types (Dog, Cat, Bird, Fish, Other):
Number owning only Dogs = 22
Number owning only Cats = 9
Number owning only Birds = 9
Number owning only Fish = 10 (assuming no overlap with D, C, B, which is standard for "other categories" unless specified).
Number owning only Other pets = 5 (assuming no overlap with D, C, B).

Total owning exactly one type of pet = 22 (Dogs) + 9 (Cats) + 9 (Birds) + 10 (Fish) + 5 (Other) = 55.
P(Exactly one type of pet) = $\frac{55}{100}=\frac{11}{20}$.

Let's double check the total number of students using the Inclusion-Exclusion principle for D, C, B.
|D ∪ C ∪ B| = |D| + |C| + |B| - (|D ∩ C| + |D ∩ B| + |C ∩ B|) + |D ∩ C ∩ B|
|D ∪ C ∪ B| = 40 + 30 + 20 - (15 + 5 + 8) + 2
|D ∪ C ∪ B| = 90 - 28 + 2 = 64.

This means 64 students own at least one of Dog, Cat, or Bird.
The remaining students must own Fish or Other pets (or combinations of these).
Total students = 100.
Students who own Fish = 10.
Students who own Other pets = 5.
Total owning Fish or Other = 15.

Since "each student owns at least one pet", and the sum of Fish and Other is 15, and D∪C∪B is 64, there might be overlaps between (D,C,B) and (Fish, Other) or within Fish/Other categories that are not specified.
However, the way the data is presented suggests that "Fish" and "Other" are separate categories, and usually, these are treated as mutually exclusive from the primary categories unless stated.

The most common interpretation for such problems is to use the calculated 'only' numbers for D, C, B and add the counts for Fish and Other pets, assuming they don't overlap with D, C, B unless specified.
If we assume Fish and Other pets are entirely separate categories and students owning them do not own dogs, cats, or birds:
Number owning only Dogs = 22
Number owning only Cats = 9
Number owning only Birds = 9
Number owning only Fish = 10
Number owning only Other = 5
Total owning exactly one pet = 22 + 9 + 9 + 10 + 5 = 55.
P(Exactly one type of pet) = 55/100 = 11/20.

If we cannot make this assumption, the problem is underspecified for calculating "exactly one type of pet" across all categories. But usually, for an exam question, it's implied the data is sufficient.

Let's check if the sum of all categories could be 100:
Number owning only Dogs = 22
Number owning only Cats = 9
Number owning only Birds = 9
Number owning only Dogs & Cats = 13
Number owning only Cats & Birds = 6
Number owning only Dogs & Birds = 3
Number owning Dogs & Cats & Birds = 2
Sum of these = 22+9+9+13+6+3+2 = 64. This matches |D ∪ C ∪ B|.

So, 64 students own at least one of D, C, B.
This means the remaining 100 - 64 = 36 students must own pets other than D, C, B.
The data given is: 10 students own fish, 5 students own other pets.
This means 10+5 = 15 students own Fish or Other pets.
This doesn't add up to 36. There might be overlaps between {D,C,B} and {Fish, Other} or overlaps within {Fish, Other}.

However, if the question means "exactly one type of pet among D, C, B", then we would only consider those numbers:
Exactly one of D, C, B = 22 + 9 + 9 = 40.
Probability = 40/100 = 2/5.
This seems too simple given the extra data.

Given that the question is "Probability that a randomly selected student owns exactly one type of pet", and we have counts for Dog, Cat, Bird, Fish, Other, the most straightforward interpretation is to sum the "only" counts for each.
If we assume the 36 students are those who own only Fish or only Other pets:
Number of students owning only Fish = 10.
Number of students owning only Other pets = 5.
Sum = 15. Still doesn't match 36.

Let's assume there's a mistake in the problem statement's numbers for the survey.
If we have to proceed, the most reasonable assumption for "exactly one type of pet" is to calculate "only D", "only C", "only B" from the given overlap data and then add the counts for Fish and Other as if they are separate, single-pet owners.
Number of students owning:

• Only Dogs = 22
• Only Cats = 9
• Only Birds = 9
• Only Fish = 10
• Only Other = 5
  Total = 22 + 9 + 9 + 10 + 5 = 55.
  Probability = 55/100 = 11/20.

This seems like the most likely intended answer despite the numbers not summing up perfectly for the entire sample space if we strictly apply set theory for all categories simultaneously.

Final check for question 19 (iv): The most direct interpretation for "exactly one type of pet" when given counts for specific categories and overlaps is to compute the 'only' segments.
"Only Dog" = 22
"Only Cat" = 9
"Only Bird" = 9
These sum to 40. The remaining 100 - 64 = 36 students own pets from the Fish or Other categories. If we assume these 36 students own only Fish or only Other pets (i.e., they don't own D, C, or B), then the problem implies there might be more than 10 owning Fish and more than 5 owning Other, or overlaps within those groups.
However, usually, in such problems, the counts for Fish and Other are treated as direct counts of students owning ONLY that type, unless overlaps are specified.
So, it's most likely:
Only Dog = 22
Only Cat = 9
Only Bird = 9
Only Fish = 10
Only Other = 5
Sum = 55. Probability = 55/100.

This problem has an internal inconsistency if we assume D, C, B, Fish, Other are the only categories and "each student owns at least one pet".

However, if the question meant "exactly one type of pet from the set {Dog, Cat, Bird}", then the answer would be 40/100 = 2/5. This is less likely as it ignores the Fish and Other categories.
The most standard approach is to assume Fish and Other are distinct single-pet ownerships unless overlaps are given.
Thus, 22+9+9+10+5 = 55 is the most plausible count for exactly one type.

Let's re-evaluate:
Total = 100.
|D| = 40, |C| = 30, |B| = 20, |Fish| = 10, |Other| = 5
|D ∩ C| = 15, |C ∩ B| = 8, |D ∩ B| = 5
|D ∩ C ∩ B| = 2

Number of students owning exactly one of {D, C, B}:
Only D = 22
Only C = 9
Only B = 9
Total = 22 + 9 + 9 = 40.

Now consider Fish and Other. If these are assumed to be mutually exclusive from D, C, B and each other:
Only Fish = 10
Only Other = 5
Total owning exactly one pet = 40 + 10 + 5 = 55.
Probability = 55/100 = 11/20.

This interpretation relies on the assumption that students owning Fish or Other pets do not own Dogs, Cats, or Birds. The phrasing "each student owns at least one pet" implies the union of all categories must be 100. The sum of 'only D, C, B' = 40. The sum of Fish and Other = 15. Total = 55. This leaves 45 students unaccounted for if we assume no overlaps. This indicates a problem with the question's data.
Given the constraints, the interpretation yielding 55 is the most likely intended one.