## Probability

Question 1 |

Let A and B be any two arbitrary events, then, which one of the following is true?

P(A∩B) = P(A)P(B) | |

P(A∪B) = P(A) + P(B) | |

P(A|B) = P(A∩B)P(B) | |

P(A∪B) ≤ P(A) + P(B) |

(B) Happens when A and B are mutually exclusive.

(C) Not happens.

(D) P(A∪B) ≤ P(A) + P(B) is true because P(A∪B) = P(A) + P(B) - P(A∩B).

Question 2 |

The probability of an event B is P_{1}. The probability that events A and B occur together is P_{2} while the probability that A and occur together is P_{3}. The probability of the event A in terms of P_{1}, P_{2} and P_{3} is __________.

P _{2} + P_{3} |

P

_{3}= P(A) - P

_{2}

P(A) = P

_{2}+ P

_{3}

Question 3 |

Let A, B and C be independent events which occur with probabilities 0.8, 0.5 and 0.3 respectively. The probability of occurrence of at least one of the event is __________

0.93 |

Since all the events are independent, so we can write

P(A∪B∪C) = P(A) + P(B) + P(C) - P(A)P(B) - P(B)P(C) - P(A)P(C) + P(A)P(B) P(C)

= 0.8 + 0.5 + 0.3 - 0.4 - 0.5 - 0.24 + 0.12

= 0.93

Question 4 |

The probability that a number selected at random between 100 and 999 (both inclusive) will not contain the digit 7 is:

16/25 | |

(9/10) ^{3} | |

27/75 | |

18/25 |

Question 5 |

A bag contains 10 white balls and 15 black balls. Two balls are drawn in succession. The probability that one of them is black and the other is white is:

2/3 | |

4/5 | |

1/2 | |

2/1 |

Probability of first ball black and second one white is,

Question 6 |

For n>2, let a ∈ {0,1}^{n} be a non-zero vector. Suppose that x is chosen uniformly at random from {0,1}^{n}. Then, the probability that is an odd number is _____.

0.5 |

^{n}

‘x’ is a vector chosen randomly from {0,1}

^{n}

‘a’ can have 2(

^{n}-1) possibilities, x can have 2

^{n}possibilities.

∑a

_{i}x

_{i}have (2

^{n}-1)(2

^{n}) possibilities, which is an even number of outcomes.

The probability of https://solutionsadda.in/wp-content/uploads/2020/02/41.jpg is odd is ½.

For example:

Take n=3

a = {001, 010, 100, 011, 101, 111}

x = {000, 001, 010, 011, 100, 101, 110, 111}

Computed as [001]×[000] = 0+0+0 = 0 Output = even

[001]×[001] = 0+0+1 = 0 Output = odd

Similarly, there could be 28 even, 28 odd outputs for the a(size=7), x(size=8) of total 56 outputs.

Question 7 |

Consider a random variable X that takes values +1 and −1 with probability 0.5 each. The values of the cumulative distribution function F(x) at x = −1 and +1 are

0 and 0.5 | |

0 and 1 | |

0.5 and 1 | |

0.25 and 0.75 |

The sum of probabilities at x=1, x=-1 itself is 0.5+0.5 = 1. It is evident that, there is no probability for any other values.

The F(x=-1) is 0.5 as per given probabilities and

F(x=1) = sum of F(x=-1) +F(x=0) = ...f(X=1) = 0.5 +0.5 = 1

Question 8 |

Suppose a fair six-sided die is rolled once. If the value on the die is 1, 2, or 3, the die is rolled a second time. What is the probability that the sum total of values that turn up is at least 6?

10/21 | |

5/12 | |

2/3 | |

1/6 |

The value on second time can be {1, 2, 3, 4, 5, 6}

So the Sum can be

We have Sample space = 36

The no. of events where (Sum = atleast 6) = {6, 7, 6, 7, 8, 6, 7, 8, 9}

So the probability atleast ‘6’ while getting {1, 2, 3} in first time = 9/36 → ①

If we get ‘6’ in the first time itself, then we do not go for rolling die again.

So, its probability = 1/6

Total probability = 1/6 + 9/36 = 1/6 + 1/4 = 10/24 = 5/12

Question 9 |

Two dice are thrown simultaneously. The probability that at least one of them will have 6 facing up is

1/36 | |

1/3 | |

25/36 | |

11/36 |

1 - (5/6 × 5/6) = 1 - (25/36) = 11/36

Question 10 |

The probability that top and bottom cards of a randomly shuffled deck are both aces in

4/52×4/52 | |

4/52×3/52 | |

4/52×3/51 | |

4/52×4/51 |

_{1}: First card being ace

E

_{2}: Last card being ace

Note that E

_{1}and E

_{2}are dependent events, i.e., probability of last card being ace if first is ace will be lesser than the probability of last card being ace if first card is not ace.

So, probability of first card being ace = 4/52

Probability of last card being ace given that first card is ace is,

P(E

_{2}/ E

_{1}) = 3/51

∴ P(E

_{1}and E

_{2}) = P(E

_{1}) ⋅ P(E

_{2}/ E

_{1}) = 4/52 × 3/51

Question 11 |

The probability that it will rain today is 0.5. The probability that it will rain tomorrow is 0.6. The probability that it will rain either today or tomorrow is 0.7. What is the probability that it will rain today and tomorrow?

0.3 | |

0.25 | |

0.35 | |

0.4 |

P(Tomorrow) = 0.6

P(T∪To) = 0.7

P(T∩To) = P(T) + P(To) - P(T∪To)

= 0.5 + 0.6 - 0.07

= 1.1 - 0.7

= 0.4

Question 12 |

A die is rolled three times. The probability that exactly one odd number turns up among the three outcomes is

1/6 | |

3/8 | |

1/8 | |

1/2 |

Total no. of possibilities are 8.

Probability of getting exactly one odd = 3/8

Question 13 |

There is a sample point at which X has the value 5. | |

There is a sample point at which X has value greater than 5. | |

There is a sample point at which X has a value greater than or equal to 5. | |

None of the above |

E(X) = x

_{1}P

_{1}+ x

_{2}P

_{2}+ ... + x

_{n}P

_{n}

In question, E(X) is given as 5.

E(X) = 5, 0≤P

_{i}≤1

P

_{1}+ P

_{2}+ ... + P

_{n}= 1 [Probability]

Therefore, E(X) = 5 is possible only if atleast one of the x

_{i}value is greater than 5.

Question 14 |

Consider two events E_{1} and E_{2} such that probability of E_{1}, Pr[E_{1}]=1/2, probability of E_{2}, Pr[E_{2}]=1/3, and probability of E_{1} and E_{2}, Pr[E_{1} and E_{2}]=1/5. Which of the following statement is/are True?

Pr[E _{1} or E_{2}] is 2/3 | |

Events E _{1} and E_{2} are independent | |

Events E _{1} and E_{2} are not independent | |

Pr[E _{1}/E_{2}] = 4/5 |

_{1}and E

_{2}are independent

then P(E

_{1}and E

_{2}) = P(E

_{1}) × P(E

_{2})

But in the given equation,

P(E

_{1}and E

_{2}) = 1/5

P(E

_{1}) × P(E

_{2}) = 1/2 × 1/3 = 1/6

So, clearly Events E

_{1}and E

_{2}are not independent.

Question 15 |

E_{1} and E_{2} are events in a probability space satisfying the following constraints:

- • Pr(E

_{1}) = Pr(E

_{2})

• Pr(E

_{1}U E

_{2}) = 1

• E

_{1}and E

_{2}are independent

The value of Pr(E_{1}), the probability of the event E_{1} is

0 | |

1/4 | |

1/2 | |

1 |

_{1}) = Pr(E

_{2}) = a(say)

Pr(E

_{1}∪ E

_{2}) = 1

E

_{1}and E

_{2}are Independent.

P(E

_{1}∪ E

_{2}) = Pr(E

_{1}) + Pr(E

_{2}) - Pr(E

_{1}∩ E

_{2})

1 = a + a - Pr(E

_{1})⋅Pr(E

_{2})

1 = a + a - a⋅a

2a - a

^{2}= 1

a

^{2}- 2a + 1 = 0

(a - 1)

^{2}= 0

a = 1

Pr(E

_{1}) = 1

Answer: Option (D)

Question 16 |

Seven (distinct) car accidents occurred in a week. What is the probability that they all occurred on the same day?

1/7 ^{7} | |

1/7 ^{6} | |

1/2 ^{7} | |

7/2 ^{7} |

^{7}

Such as total no. of days in a week = 7

Total probability = 7 × 1/7

^{7}= 1/7

^{6}

Question 17 |

Four fair coins are tossed simultaneously. The probability that at least one head and one tail turn up is

1/16 | |

1/8 | |

7/8 | |

15/16 |

^{2}= 16

Atleast one head = 1/16

Atleast one tail = 1/16

Probability of getting one head and one tail is = 1 - 1/16 - 1/16 = 16 - 1 - 1/16 = 14/16 = 7/8

Question 18 |

Let P(E) denote the probability of the event E. Given P(A) = 1, P(B) = 1/2, the values of P(A|B) and P(B|A) respectively are

1/4, 1/2 | |

1/2, 1/4 | |

1/2, 1 | |

1, 1/2 |

P(A/B) = P(A∩B)/P(B)

= P(A)⋅P(B)/P(B) (consider P(A), P(B) are two independent events)

= 1

P(B/A) = P(B∩A)/P(A)

= P(B)⋅P(A)/P(A)

= 1/2

Question 19 |

A program consists of two modules executed sequentially. Let f_{1}(t) and f_{2}(t) respectively denote the probability density functions of time taken to execute the two modules. The probability density function of the overall time taken to execute the program is given by:

f _{1}(t) + f_{2}(t) | |

max{f _{1}(t), f_{2}(t)} |

_{1}(t) and f

_{2}(t) are executed sequentially.

→ They representing the probability density functions of time taken to execute.

→ f

_{1}can be executed in 'x' time.

f

_{2}can be executed in 't-x' time.

→ The probability density function =

Question 20 |

Let X and Y be two exponentially distributed and independent random variables with mean α and β, respectively. If Z = min(X,Y), then the mean of Z is given by

1/α+β | |

min(α, β) | |

αβ/α + β | |

α + β |

Question 21 |

In how many ways can we distribute 5 distinct balls, B_{1}, B_{2}, …, B_{5} in 5 distinct cells, C_{1}, C_{2}, …, C_{5} such that Ball B, is not in cell C_{i}, ∀i = 1, 2, …, 5 and each cell contains exactly one ball?

44 | |

96 | |

120 | |

3125 |

∠5(1 - 1/∠1 + 1/∠2 - 1/∠3 + 1/∠4 - 1/∠5)

= 44

Question 22 |

A bag contains 10 blue marbles, 20 green marbles and 30 red marbles. A marble is drawn from the bag, its colour recorded and it is put back in the bag. This process is repeated 3 times. The probability that no two of the marbles drawn have the same colour is

1/36 | |

1/6 | |

1/4 | |

1/3 |

Total possible combinations = 3! = 6

Probability of blue marble = 10/60[10 + 20 + 30 = 60]

Probability of green marble = 20/60

Probability of red marble = 30/60

The probability that no two of the marbles has same colour = [10/60 * 20/60 * 30/60] = 1/6

Question 23 |

An unbiased coin is tossed repeatedly until the outcome of two successive tosses is the same. Assuming that the trials are independent, the expected number of tosses is

3 | |

4 | |

5 | |

6 |

The expected number of coin flips for getting n consecutive heads is(2^(n+1) -2)

The expected number of coin flips for getting n consecutive heads is(2^(n+1) -2)

The expected number of coin flips for getting n consecutive same tosses is (2^(n+1) -2) / 2.

where n = 2,

which is (2^(3+1) - 2) / 2 = 3

Question 24 |

In a certain town, the probability that it will rain in the afternoon is known to be 0.6. Moreover, meteorological data indicates that if the temperature at noon is less than or equal to 25°C, the probability that it will rain in the afternoon is 0.4. The temperature at noon is equally likely to be above 25°C, or at/below 25°C. What is the probability that it will rain in the afternoon on a day when the temperature at noon is above 25°C?

0.4 | |

0.6 | |

0.8 | |

0.9 |

0.6 = (0.5×0.4) + (0.5×P(rain at temp>25)

0.6 = (2) + (0.5×P(rain at temp>25)

P(rain at temp>25) = 0.8

Question 25 |

When a coin is tossed, the probability of getting a Head is p, 0 < p < 1. Let N be the random variable denoting the number of tosses till the first Head appears, including the toss where the Head appears. Assuming that successive tosses are independent, the expected value of N is

1/p | |

1/(1-p) | |

1/p ^{2} | |

1/(1-p ^{2}) |

Multiply both sides with (1 - p) and subtract,

E - (1 - p) E = 1 × p + (1 - p) p + (1 - p) (1 - p) p + ......

E - (1 - p) E = p/(1 - (1 - p))

(1 - 1 + p) E = 1

pE = 1

E = 1/p

Question 26 |

Suppose there are two coins. The first coin gives heads with probability 5/8 when tossed, while the second coin gives heads with probability 1/4. One of the two coins is picked up at random with equal probability and tossed. What is the probability of obtaining heads ?

7/8 | |

1/2 | |

7/16 | |

5/32 |

The probability of obtaining heads

= (1/2)(5/8) + (1/2)(1/4)

= (5/16) + (1/8)

= 7/16

Question 27 |

Consider a hash function that distributes keys uniformly. The hash table size is 20. After hashing of how many keys will the probability that any new key hashed collides with an existing one exceed 0.5.

5 | |

6 | |

7 | |

10 |

Probability of collision for each entry = 1/20

After inserting X values then probability becomes 1/2

i.e., (1/20)X = 1/2

X = b

Question 28 |

In a multi-user operating system on an average, 20 requests are made to use a particular resource per hour. The arrival of requests follows a Poisson distribution. The probability that either one, three or five requests are made in 45 minutes is given by :

6.9 × 10 ^{6} × e^{-20} | |

1.02 × 10 ^{6} × e^{-20} | |

6.9 × 10 ^{3} × e^{-20} | |

1.02 × 10 ^{3} × e^{-20} |

_{0}request in 1 hour. So we can expect 15 requests in 45 minutes.

So, λ=15

Question 29 |

A sample space has two events A and B such that probabilities P(A ∩ B) = 1/2, P(A') = 1/3, P(B') = 1/3. What is P(A ∪ B)?

11/12 | |

10/12 | |

9/12 | |

8/12 |

P(A') = 1/3; P(A) = 2/3

P(B') = 1/3; P(B) = 2/3

P(A ∪ B) = P(A) +P(B) - P(A ∩ B)

= 2/3 + 2/3 - 1/2

= 4+4-3/ 6

= 5/6

= 10/12

Question 30 |

What is the probability that in a randomly chosen group of r people at least three people have the same birthday?

Question 31 |

Which of the following statements are FALSE?

For poisson distribution, the mean is twice the variance. | |

In queuing theory, if arrivals occur according to poisson distribution, then the inter-arrival time is exponentially distributed. | |

The distribution of waiting time is independent of the service discipline used in selecting the waiting customers for service. | |

If the time between successive arrivals is exponential, then the time between the occurrences of every third arrival is also exponential. | |

Both (A) and (C). |

Option C is also False, because waiting time is dependent on the service discipline.

Question 32 |

If a fair coin is tossed four times. What is the probability that two heads and two tails will result?

3/8 | |

1/2 | |

5/8 | |

3/4 |

Then total number of possibilities = 2

^{4}= 16

No. of possibilities getting 2 heads and 2 tails is

HHTT, HTHT, TTHH, THTH, THHT, HTTH = 6

Probability of getting 2 heads and 2 tails is

= No. of possibilities/Total no. of possibilities = 6/16 = 3/8

Question 33 |

An examination paper has 150 multiple-choice questions of one mark each, with each question having four choices. Each incorrect answer fetches -0.25 mark. Suppose 1000 students choose all their answers randomly with uniform probability. The sum total of the expected marks obtained by all these students is:

0 | |

2550 | |

7525 | |

9375 |

Probability of selecting a wrong answer = 3/4

For correct answer +1, for wrong answer-0.25;

Expected marks for each question = (1/4) × 1 + (3/4) -(0.25)

= 1/4 + (-3/16)

= 4-3/16

= 1/16

= 0.0625

Expected marks for 150 questions = 150 × 0.625 = 9.375

The sum total of expected marks obtained by 1000 students is = 1000×9.375 = 9375

Question 34 |

Two n bit binary strings, S_{1} and S_{2}, are chosen randomly with uniform probability. The probability that the Hamming distance between these strings (the number of bit positions where the two strings differ) is equal to d is

^{n}C_{d} /2^{n}
| |

^{n}C_{d} / 2^{d} | |

d/2 ^{n} | |

1/2 ^{d} |

^{n}C

_{d}

Total no. of cases where n positions have any binary bit = 2

^{n}

The probability of 'd' bits differ =

^{n}C

_{d}/ 2

^{n}

Question 35 |

A point is randomly selected with uniform probability in the X-Y plane within the rectangle with corners at (0,0), (1,0), (1,2) and (0,2). If p is the length of the position vector of the point, the expected value of p^{2} is

2/3 | |

1 | |

4/3 | |

5/3 |

Above diagram shows the scenario of our question.

The length p of our position vector (x,y) is

p = √x

^{2}+ y

^{2}

p

^{2}= x

^{2}+ y

^{2}

E(p

^{2}) = E(x

^{2}+ y

^{2}) = E(x

^{2}) + E(y

^{2})

Now we need to calculate the probability density function of X and Y.

Since distribution is uniform,

X goes from 0 to 1, so PDF(x) = 1/1-0 = 1

Y goes from 0 to 2, so PDF(y) = 1/2-0 = 1/2

Now we evaluate,

E(p

^{2}) = E(x

^{2}) + E(y

^{2}) = 5/3

Question 36 |

Let S be a sample space and two mutually exclusive events A and B be such that A∪B = S. If P(∙) denotes the probability of the event, the maximum value of P(A)P(B) is __________.

0.25 | |

0.26 | |

0.27 | |

0.28 |

P(A∪B) = P(A) + P(B) + P(A∩B) = 1 →①

But, as A and B are mutually exclusive events

P(A∩B) = 0

∴ P(A∪B) = P(A) + P(B) = 1 →②

Arithmetic mean of two numbers ≥ Geometric mean of those two numbers

(P(A)+P(B))/2 ≥ √(P(A)∙P(B))

1/2 ≥ √(P(A)∙P(B)) (∵from ②)

Squaring on both sides

1/4 ≥ P(A)∙P(B)

P(A)∙P(B) ≤ 1/4

∴ Maximum value of P(A)P(B) = 1/4 = 0.25