# Probability Questions for IIFT PDF

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## Probability Questions for IIFT PDF

Download important IIFT Probability Questions PDF based on previously asked questions in IIFT and other MBA exams. Practice Probability questions and answers for IIFT exam.

Practice IIFT Mock Tests

Question 1: Ramesh plans to order a birthday gift for his friend from an online retailer. However, the birthday coincides with the festival season during which there is a huge demand for buying online goods and hence deliveries are often delayed. He estimates that the probability of receiving the gift, in time, from the retailers A, B, C and D would be 0.6, 0.8, 0.9 and 0.5 respectively.

Playing safe, he orders from all four retailers simultaneously. What would be the probability that his friend would receive the gift in time?

a) 0.004

b) 0.006

c) 0.216

d) 0.994

e) 0.996

Question 2: The probability that a randomly chosen positive divisor of $10^{29}$ is an integer multiple of $10^{23}$ is: $a^{2} /b^{2}$, then ‘b – a’ would be:

a) 8

b) 15

c) 21

d) 23

e) 45

Question 3: Aditya has a total of 18 red and blue marbles in two bags (each bag has marbles of both colors). A
marble is randomly drawn from the first bag followed by another randomly drawn from the
second bag, the probability of both being red is 5/16. What is the probability of both marbles being blue?

a) 1/16

b) 2/16

c) 3/16

d) 4/16

e) 5/16

Question 4: A bag contains 8 red and 6 blue balls. If 5 balls are drawn at random, what is the probability that 3 of them are red and 2 are blue?

a) 80/143

b) 50/143

c) 75/143

d) None of the above

Question 5: The answer sheets of 5 engineering students can be checked by any one of 9 professors. What is the probability that all the 5 answer sheets are checked by exactly 2 professors?

a) 20/2187

b) 40/2187

c) 40/729

d) None of the above

Question 6: The probability that in a household LPG will last 60 days or more is 0.8 and that it will last at most 90 days is 0.6. The probability that the LPG will last 60 to 90 days is

a) 0.40

b) 0.50

c) 0.75

d) None of the above

Question 7: Two trains P and Q are scheduled to reach New Delhi railway station at 10.00 AM. The probability that train P and train Q will be late is 7/9 and 11/27 respectively. The probability that train Q will be late, given that train P is late, is 8/9. Then the probability that neither train will be late on a particular day is

a) 40/81

b) 41/81

c) 77/81

d) 77/243

Question 8: A dice is rolled twice. What is the probability that the number in the second roll will be higher than that in the first?

a) 5/36

b) 8/36

c) 15/36

d) 21/36

e) None of the above

Question 9: Six playing cards are lying face down on a table, two of them are kings. Two cards are drawn at random. Let a denote the probability that at least one of the cards drawn is a king, and b denote the probability of not drawing a king. The ratio a/b is

a) $\geq0.25$and$<0.5$

b) $\geq0.5$and$<0.75$

c) $\geq0.75$and$<1.0$

d) $\geq1.0$and$<1.25$

e) $\geq1.25$

Question 10: A coin of radius 3 cm is randomly dropped on a square floor full of square shaped tiles of side 10 cm each. What is the probability that the coin will land completely within a tile? In other words, the coin should not cross the edge of any tile.

a) 0.91

b) 0.5

c) 0.49

d) 0.36

e) 0.16

That can be calculated as the probability of his friend receiving at least one gift.

The probability that none of the retailers sends in time = $(1 – 0.6) \times (1 – 0.8) \times (1 – 0.9) \times (1 – 0.5)$

= $0.4 \times 0.2 \times 0.1 \times 0.5 = 0.004$

$\therefore$ Probability of his receiving at least one gift = $1 – 0.004 = 0.996$

Number of factors of $10^{29} = 2^{29} \times 5^{29}$

= $30 \times 30 = 900$

Factors of $10^{29}$ which are multiple of $10^{23}$

= $10^6 = 2^6 \times 5^6$

= $7 \times 7 = 49$

=> Required probability = $\frac{49}{900} = \frac{a^2}{b^2}$

=> $\frac{a}{b} = \frac{7}{30}$

$\therefore b – a = 30 – 7 = 23$

Probability of both marbles being red = Probability of red from 1st * probability of red from 2nd

= $\frac{5}{16}$

$\because$ Each bag has marbles of both colors and probability cannot be greater than 1

=> $\frac{5}{16} = \frac{5}{8} \times \frac{1}{2}$

where, probability of red marbles from 1st bag = $\frac{5}{8}$

=> Probability of blue marbles from 1st bag = $1 – \frac{5}{8} = \frac{3}{8}$

Similarly, Probability of red marbles from 1st bag = $\frac{1}{2}$

=> Probability of blue marbles from 2nd bag = $1 – \frac{1}{2} = \frac{1}{2}$

$\therefore$ Probability of both blue marbles = $\frac{3}{8} \times \frac{1}{2}$

= $\frac{3}{16}$

5 balls are drawn at random, hence the probability that 3 of them are red and 2 are blue = $\dfrac{8c3*6c2}{14c5}$ = $\dfrac{60}{143}$

Hence, option D is the correct answer.

Each of the 5 papers can be checked by any of the 9 professors and thus the total number of way = 9$^5$
Selecting exactly 2 of the 9 professors can be done in $^9C_2$ way = 36
They can correct the 5 papers in 2$^5$ – 2(all the 5 papers checked by the same professor) = 30  ways.
Thus, the total number of ways = 36*30
Hence, the required probability = $\dfrac{36*30}{9^5}$ = $\dfrac{40}{2187}$
Hence, option B is the correct answer.

The probability that the LPG will last atmost 90 days = 0.6
Thus, the probability that the LPG will last more than 90 days = 0.4
The probability that the LPG will last more than 60 days = 0.8 = The probability that the LPG will last 60 to 90 days + The probability that the LPG will last more than 90 days
Hence, 0.4 + The probability that the LPG will last 60 to 90 days = 0.8
Hence, The probability that the LPG will last 60 to 90 days = 0.8- 0.4 = 0.4
Hence, option A is the correct answer.

Let ‘A’ and ‘B’ be the event of train reaching at the station respectively.

P(A)$_{\text{Late}}$ = $\dfrac{7}{9}$, therefore, P(A)$_{\text{On time}}$ = $\dfrac{2}{9}$.

P(B)$_{\text{Late}}$ = $\dfrac{11}{27}$, therefore, P(B)$_{\text{On time}}$ = $\dfrac{16}{27}$.

The probability that train Q will be late, given that train P is late, is 8/9.

P$(\dfrac{B_{\text{Late}}}{A_{\text{Late}}})$=$\dfrac{8}{9}$

P(A$_{\text{Late}} \cap$ B$_{\text{Late}})$ = P(A)$_{\text{Late}}$*P($\dfrac{B_{\text{Late}}}{A_{\text{Late}}})$

P(A$_{\text{Late}} \cap$ B$_{\text{Late}})$ = $\dfrac{7}{9}$*$\dfrac{8}{9} = \dfrac{56}{81}$

Therefore, the probability that neither train is late = 1 – (P(A)$_{\text{Late}}$+P(B)$_{\text{Late}}$ – P(A$_{\text{Late}} \cap$ B$_{\text{Late}})$)

$\Rightarrow$ 1 – ($\dfrac{7}{9}$+$\dfrac{11}{27}$-$\dfrac{56}{81}$)

$\Rightarrow$ $\dfrac{41}{81}$

Hence, we can say that option B is the correct answer.

A die is rolled twice.
The number of combinations that can occur = 6*6 = 36.
We have to find the probability of the second roll being higher than the first.
If we select 2 numbers out of the 6 and arrange them in ascending order, then we will obtain the scenario in which the number obtained in the second roll will be greater than the number obtained in the first roll.

2 numbers out of 6 numbers can be selected in 6C2 = 15 ways. The numbers can be arranged in ascending order in only one way.
Therefore, the required probability is 15/36.

Therefore, option C is the right answer.

There are 6 cards and 2 out of the 6 cards are kings.
Number of ways of selecting 2 cards = 6C2 = 15 ways.
Number of ways in which 2 cards can be selected such that both of them are King = 2C2 = 1
Number of ways in which 2 cards can be selected such that exactly one of them is a King = 2C1*4C1 = 8
=> a = (1+8)/15 = 9/15
b = 1-(9/15) = 6/15
a/b = 9/6 = 1.5
1.5 > 1.25
Therefore, option E is the right answer.