Probability Axioms and Conditional Probability Questions and Answers

Q: Consider a permutation sampled uniformly at random from the set of all permutations of \(\{1, 2, 3, \cdots, n\}\) for some \(n \ge 4\). Let \(X\) be the event that 1 occurs before 2 in the permutation, and \(Y\) the event that 3 occurs before 4. Which one of the following statements is TRUE?

The events \(X\) and \(Y\) are independent.By symmetry, 1 precedes 2 in half of all permutations, so \(P(X) = 1/2\), and likewise \(P(Y) = 1/2\). The relative order of 1 and 2 and the relative order of 3 and 4 involve disjoint pairs of elements, and all four combinations of orders are equally likely, so\[ P(X \cap Y) = \frac{1}{4} = P(X)\,P(Y) \]They are not mutually exclusive, since both can happen (for example 1, 2, 3, 4, ...). Neither must occur: both fail together with probability 1/4. And X is not more likely than Y, since both have probability 1/2.

Q: An urn contains one red ball and one blue ball. At each step, a ball is picked uniformly at random from the urn, and this ball together with another ball of the same color is put back in the urn. The probability that there are equal number of red and blue balls after two steps is

The probability is 1/3.After two steps the urn holds 4 balls. Equal numbers means 2 red and 2 blue, which needs one red draw and one blue draw.\[ \begin{aligned} P(R \text{ then } B) &= \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6} \\ P(B \text{ then } R) &= \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6} \end{aligned} \]After the first draw the urn holds 3 balls, two of the drawn colour, so the other colour has probability 1/3. The total is 1/6 + 1/6 = 1/3.1/2 treats the second draw as if the urn were unchanged, which ignores the reinforcement. The other outcomes, 3 red and 1 blue or the reverse, have probability 1/3 each, which is a useful check: 1/3 + 1/3 + 1/3 = 1.

Q: In a year of 366 days, what is the chance that three persons have different birthdays?

The probability is \(\frac{365 \times 364}{366 \times 366}\).Take the three people one at a time. The first can have any of the 366 days. The second must avoid that day, and the third must avoid both earlier days:\[ \begin{aligned} P &= \frac{366}{366} \times \frac{365}{366} \times \frac{364}{366} \\ &= \frac{365 \times 364}{366 \times 366} \end{aligned} \]The option \(1 - \frac{365 \times 364}{366 \times 366}\) is the complement: the chance that at least two of the three share a birthday. \(\frac{1}{366^{3}}\) is the chance that the three birthdays are three particular stated days, and \(\frac{1}{366 \times 365 \times 364}\) does not correspond to the event asked about; both are far too small.

Q: 2 fair coins are tossed. What is the probability of getting exactly 2 heads or 2 tails ?

The probability is \(\frac{1}{2}\).Two fair coins give four equally likely outcomes: HH, HT, TH and TT. Exactly 2 heads is HH, and exactly 2 tails is TT; the two events cannot happen together, so their probabilities add:\[ P = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \]\(\frac{1}{4}\) is the probability of only one of the two events, for example just HH. \(\frac{1}{3}\) comes from wrongly treating HH, TT and 'one of each' as equally likely, and \(\frac{3}{8}\) belongs to a three-coin question.

Q: In case of n attributes the total number of ultimate class frequencies is :

With n attributes there are \(2^{n}\) ultimate class frequencies.An ultimate class is specified by stating, for every one of the n attributes, whether it is present or absent, as in ABC, ABγ, AβC and so on. Each attribute gives two choices, so the number of ultimate classes is\[ 2 \times 2 \times \cdots \times 2 = 2^{n} \]For two attributes there are four: (AB), (Aβ), (αB) and (αβ). The total number of all class frequencies, of every order, is \(3^{n}\). \(n\) and \(2n\) are far too few, and \(4^{n}\) too many.

Q: In a series of 100 houses actually invaded by smallpox, 70% of the inhabitants are attacked and 85% have been vaccinated. What is the lowest percentage of the vaccinated that must have been attacked ?

At least 64.7% of the vaccinated must have been attacked.Of the inhabitants, 70% were attacked and 85% vaccinated. The overlap is smallest when the two groups overlap as little as possible:\[ \begin{aligned} \text{Attacked and vaccinated} &\ge 70 + 85 - 100 \\ &= 55\% \end{aligned} \]As a percentage of the vaccinated:\[ \frac{55}{85} \times 100 \approx 64.7\% \]63.7%, 62.7% and 61.7% are lower than this minimum, so they are not possible for the data given.

Q: A committee of 4 people is to be appointed from 3 officers of the production department, 4 officers of the purchase department, 2 officers of the sales department and one chartered accountant. What is the probability of forming the committee that there must be one from each category ?

The probability is \(\frac{8}{70}\).There are \(3 + 4 + 2 + 1 = 10\) people. One from each of the four categories can be chosen in \(3 \times 4 \times 2 \times 1\) ways.\[ \begin{aligned} \text{total} &= \binom{10}{4} = 210 \\ \text{favourable} &= 3 \times 4 \times 2 \times 1 = 24 \\ P &= \frac{24}{210} = \frac{8}{70} \end{aligned} \]\(\frac{8}{70}\) is \(\frac{4}{35}\), left unreduced to match the options. The other options correspond to 6, 12 and 18 favourable committees out of 210, all undercounts of the 24 ways to pick one person from each category.

Q: A deck of playing cards contains 52 cards, 4 each with face values in the set {A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K}. Let us define face values of the top and bottom cards as X and Y. The shuffling algorithm is defined as follows :Sh1 : Permute the cards randomly due to that each arrangement occurs with probability 1/52 !Sh2 : If \(X \ne Y\), flip the biased coin that comes up heads with probability P and go back to Sh1 if heads turns up otherwise stop.Each coin flip and each permutation is assumed to be independent of all the other randomization. What value of P will make X and Y independent random variables once this process of shuffling will stops ?

The required value is P = 1/4.After Sh1, a specific pair of face values for (top, bottom) has probability depending on whether the values match. The process keeps a matching pair at once, but keeps a non-matching pair only when the coin shows tails, with probability 1 - P. So the final probabilities are proportional to\[ \begin{aligned} (x, x) &: \tfrac{4}{52} \cdot \tfrac{3}{51} = \tfrac{12}{2652} \\ (x, y) &: \tfrac{4}{52} \cdot \tfrac{4}{51}(1 - P) \\ &= \tfrac{16(1-P)}{2652} \end{aligned} \]X and Y are independent, each uniform over 13 values, exactly when every one of the 169 pairs is equally likely:\[ \begin{aligned} 12 &= 16(1 - P) \\ 1 - P &= \tfrac{3}{4} \\ P &= \tfrac{1}{4} \end{aligned} \]With P = 1/3, 2/3 or 1/5 the matching pairs would be over- or under-weighted, so X and Y would stay dependent.

Q: If two dice are thrown, what is the probability that the sum is neither 7 nor 11 ?

The probability is \(\frac{7}{9}\).With two dice there are 36 equally likely outcomes. A sum of 7 occurs in 6 ways and a sum of 11 in 2 ways (5-6 and 6-5), so 8 outcomes are excluded:\[ \begin{aligned} P &= 1 - \frac{6 + 2}{36} \\ &= \frac{28}{36} = \frac{7}{9} \end{aligned} \]The chance that the sum is 7 or 11 is \(\frac{8}{36} = \frac{2}{9}\); the question asks for the complement. \(\frac{1}{9}\), \(\frac{3}{9}\) and \(\frac{5}{9}\) do not match either count, so only \(\frac{7}{9}\) fits.

Consider a permutation sampled uniformly at random from the set of all permutations of \(\{1, 2, 3, \cdots, n\}\) for some \(n \ge 4\). Let \(X\) be the event that 1 occurs before 2 in the permutation, and \(Y\) the event that 3 occurs before 4. Which one of the following statements is TRUE?

A.The events \(X\) and \(Y\) are mutually exclusive
B.The events \(X\) and \(Y\) are independent
C.Either event \(X\) or \(Y\) must occur
D.Event \(X\) is more likely than event \(Y\)

Exam Year(s)

GATE CS - 2024,

Answer: B

The events \(X\) and \(Y\) are independent.

By symmetry, 1 precedes 2 in half of all permutations, so \(P(X) = 1/2\), and likewise \(P(Y) = 1/2\). The relative order of 1 and 2 and the relative order of 3 and 4 involve disjoint pairs of elements, and all four combinations of orders are equally likely, so

\[ P(X \cap Y) = \frac{1}{4} = P(X)\,P(Y) \]

They are not mutually exclusive, since both can happen (for example 1, 2, 3, 4, ...). Neither must occur: both fail together with probability 1/4. And X is not more likely than Y, since both have probability 1/2.

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An urn contains one red ball and one blue ball. At each step, a ball is picked uniformly at random from the urn, and this ball together with another ball of the same color is put back in the urn. The probability that there are equal number of red and blue balls after two steps is

A.1/4
B.1/3
C.1/2
D.2/3

Exam Year(s)

GATE CS - 2026,

Answer: B

The probability is 1/3.

After two steps the urn holds 4 balls. Equal numbers means 2 red and 2 blue, which needs one red draw and one blue draw.

\[ \begin{aligned} P(R \text{ then } B) &= \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6} \\ P(B \text{ then } R) &= \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6} \end{aligned} \]

After the first draw the urn holds 3 balls, two of the drawn colour, so the other colour has probability 1/3. The total is 1/6 + 1/6 = 1/3.

1/2 treats the second draw as if the urn were unchanged, which ignores the reinforcement. The other outcomes, 3 red and 1 blue or the reverse, have probability 1/3 each, which is a useful check: 1/3 + 1/3 + 1/3 = 1.

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In a year of 366 days, what is the chance that three persons have different birthdays?

A.\(\frac{1}{366 \times 366 \times 366}\)
B.\(\frac{1}{366 \times 365 \times 364}\)
C.\(\frac{365 \times 364}{366 \times 366}\)
D.\(1 - \frac{365 \times 364}{366 \times 366}\)

Exam Year(s)

IIT JAM - 2026,

Answer: C

The probability is \(\frac{365 \times 364}{366 \times 366}\).

Take the three people one at a time. The first can have any of the 366 days. The second must avoid that day, and the third must avoid both earlier days:

\[ \begin{aligned} P &= \frac{366}{366} \times \frac{365}{366} \times \frac{364}{366} \\ &= \frac{365 \times 364}{366 \times 366} \end{aligned} \]

The option \(1 - \frac{365 \times 364}{366 \times 366}\) is the complement: the chance that at least two of the three share a birthday. \(\frac{1}{366^{3}}\) is the chance that the three birthdays are three particular stated days, and \(\frac{1}{366 \times 365 \times 364}\) does not correspond to the event asked about; both are far too small.

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2 fair coins are tossed. What is the probability of getting exactly 2 heads or 2 tails ?

A.\(\frac{1}{4}\)
B.\(\frac{3}{8}\)
C.\(\frac{1}{2}\)
D.\(\frac{1}{3}\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

The probability is \(\frac{1}{2}\).

Two fair coins give four equally likely outcomes: HH, HT, TH and TT. Exactly 2 heads is HH, and exactly 2 tails is TT; the two events cannot happen together, so their probabilities add:

\[ P = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \]

\(\frac{1}{4}\) is the probability of only one of the two events, for example just HH. \(\frac{1}{3}\) comes from wrongly treating HH, TT and 'one of each' as equally likely, and \(\frac{3}{8}\) belongs to a three-coin question.

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In case of n attributes the total number of ultimate class frequencies is :

A.\(n\)
B.\(2n\)
C.\(2^{n}\)
D.\(4^{n}\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

With n attributes there are \(2^{n}\) ultimate class frequencies.

An ultimate class is specified by stating, for every one of the n attributes, whether it is present or absent, as in ABC, ABγ, AβC and so on. Each attribute gives two choices, so the number of ultimate classes is

\[ 2 \times 2 \times \cdots \times 2 = 2^{n} \]

For two attributes there are four: (AB), (Aβ), (αB) and (αβ). The total number of all class frequencies, of every order, is \(3^{n}\). \(n\) and \(2n\) are far too few, and \(4^{n}\) too many.

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In a series of 100 houses actually invaded by smallpox, 70% of the inhabitants are attacked and 85% have been vaccinated. What is the lowest percentage of the vaccinated that must have been attacked ?

A.64.7%
B.63.7%
C.62.7%
D.61.7%

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

At least 64.7% of the vaccinated must have been attacked.

Of the inhabitants, 70% were attacked and 85% vaccinated. The overlap is smallest when the two groups overlap as little as possible:

\[ \begin{aligned} \text{Attacked and vaccinated} &\ge 70 + 85 - 100 \\ &= 55\% \end{aligned} \]

As a percentage of the vaccinated:

\[ \frac{55}{85} \times 100 \approx 64.7\% \]

63.7%, 62.7% and 61.7% are lower than this minimum, so they are not possible for the data given.

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A committee of 4 people is to be appointed from 3 officers of the production department, 4 officers of the purchase department, 2 officers of the sales department and one chartered accountant. What is the probability of forming the committee that there must be one from each category ?

A.\(\frac{2}{70}\)
B.\(\frac{4}{70}\)
C.\(\frac{6}{70}\)
D.\(\frac{8}{70}\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: D

The probability is \(\frac{8}{70}\).

There are \(3 + 4 + 2 + 1 = 10\) people. One from each of the four categories can be chosen in \(3 \times 4 \times 2 \times 1\) ways.

\[ \begin{aligned} \text{total} &= \binom{10}{4} = 210 \\ \text{favourable} &= 3 \times 4 \times 2 \times 1 = 24 \\ P &= \frac{24}{210} = \frac{8}{70} \end{aligned} \]

\(\frac{8}{70}\) is \(\frac{4}{35}\), left unreduced to match the options. The other options correspond to 6, 12 and 18 favourable committees out of 210, all undercounts of the 24 ways to pick one person from each category.

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A deck of playing cards contains 52 cards, 4 each with face values in the set {A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K}. Let us define face values of the top and bottom cards as X and Y. The shuffling algorithm is defined as follows :

Sh1 : Permute the cards randomly due to that each arrangement occurs with probability 1/52 !
Sh2 : If \(X \ne Y\), flip the biased coin that comes up heads with probability P and go back to Sh1 if heads turns up otherwise stop.

Each coin flip and each permutation is assumed to be independent of all the other randomization. What value of P will make X and Y independent random variables once this process of shuffling will stops ?

A.P = 1/4
B.P = 2/3
C.P = 1/3
D.P = 1/5

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

The required value is P = 1/4.

After Sh1, a specific pair of face values for (top, bottom) has probability depending on whether the values match. The process keeps a matching pair at once, but keeps a non-matching pair only when the coin shows tails, with probability 1 - P. So the final probabilities are proportional to

\[ \begin{aligned} (x, x) &: \tfrac{4}{52} \cdot \tfrac{3}{51} = \tfrac{12}{2652} \\ (x, y) &: \tfrac{4}{52} \cdot \tfrac{4}{51}(1 - P) \\ &= \tfrac{16(1-P)}{2652} \end{aligned} \]

X and Y are independent, each uniform over 13 values, exactly when every one of the 169 pairs is equally likely:

\[ \begin{aligned} 12 &= 16(1 - P) \\ 1 - P &= \tfrac{3}{4} \\ P &= \tfrac{1}{4} \end{aligned} \]

With P = 1/3, 2/3 or 1/5 the matching pairs would be over- or under-weighted, so X and Y would stay dependent.

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If two dice are thrown, what is the probability that the sum is neither 7 nor 11 ?

A.\(\frac{3}{9}\)
B.\(\frac{5}{9}\)
C.\(\frac{7}{9}\)
D.\(\frac{1}{9}\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The probability is \(\frac{7}{9}\).

With two dice there are 36 equally likely outcomes. A sum of 7 occurs in 6 ways and a sum of 11 in 2 ways (5-6 and 6-5), so 8 outcomes are excluded:

\[ \begin{aligned} P &= 1 - \frac{6 + 2}{36} \\ &= \frac{28}{36} = \frac{7}{9} \end{aligned} \]

The chance that the sum is 7 or 11 is \(\frac{8}{36} = \frac{2}{9}\); the question asks for the complement. \(\frac{1}{9}\), \(\frac{3}{9}\) and \(\frac{5}{9}\) do not match either count, so only \(\frac{7}{9}\) fits.

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10.

Five people stand at the vertices of a pentagon, throwing frisbees to each other. They have two frisbees, initially at adjacent vertices as shown in the figure. In each time interval, each frisbee is thrown either to the left or to the right (along an edge of the pentagon) with equal probability. This process continues until one person is the target of two frisbees simultaneously, then game ends.

(Note : All throws are independent of past history)

A regular pentagon with a person at each vertex; the two frisbees are at the two ends of the bottom edge, with arrows showing the throws

Find the mean and variance of the number of pairs of throws.

A.12 and 100
B.6 and 50
C.24 and 200
D.18 and 80

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

The mean is 12 pairs of throws and the variance is 100, so the answer is 12 and 100.

Track the distance between the frisbees around the pentagon: 1 (adjacent) or 2, with 0 ending the game. Each frisbee moves one step left or right, so the gap changes by -2, 0 or +2 with probabilities 1/4, 1/2, 1/4. From distance 1 the gap stays at 1 with probability 3/4 and becomes 2 with probability 1/4; from distance 2 it ends with probability 1/4, becomes 1 with probability 1/4 and stays at 2 with probability 1/2.

Let \(E_{1}\), \(E_{2}\) be the expected numbers of throws from each distance:

\[ \begin{aligned} E_{1} &= 1 + \tfrac{3}{4}E_{1} + \tfrac{1}{4}E_{2} \\ E_{2} &= 1 + \tfrac{1}{4}E_{1} + \tfrac{1}{2}E_{2} \end{aligned} \]

\[ \begin{aligned} E_{1} &= 4 + E_{2},\quad E_{2} = 2 + \tfrac{1}{2}E_{1} \\ E_{1} &= 12,\quad E_{2} = 8 \end{aligned} \]

For the variance, the fundamental matrix of the absorbing chain is \(N = \begin{pmatrix} 8 & 4 \\ 4 & 4 \end{pmatrix}\), and the variance of the absorption time from distance 1 is the first entry of \((2N - I)t - t^{2}\) with \(t = (12, 8)\):

\[ 15(12) + 8(8) - 12^{2} = 244 - 144 = 100 \]

The other pairs do not satisfy these equations; 6 and 50 or 24 and 200 come from halving or doubling both values.

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