# CAT Level Questions on Probability

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## CAT Level Questions on Probability

Download important CAT Questions on Probability PDF based on previously asked questions in CAT exam. Practice Probability Questions PDF for CAT exam.

Question 1: The supervisor of a packaging unit of a milk plant is being pressurised to finish the job closer to the distribution time, thus giving the production staff more leeway to cater to last minute demand. He has the option of running the unit at normal speed or at 110% of normal – “fast speed”. He estimates that he will be able to run at the higher speed for 60% of the time. The packet is twice as likely to be damaged at the higher speed which would mean temporarily stopping the process. If a packet on a randomly selected packaging runs has probability of 0.112 of damage, what is the probability that the packet will not be damaged at normal speed?

a) 0.81

b) 0.93

c) 0.75

d) 0.60

e) None of the above

Question 2: 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 3: 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 4: 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 5: There are four machines in a factory. At exactly 8 pm, when the mechanic is about to leave the factory, he is informed that two of the four machines are not working properly. The mechanic is in a hurry, and decides that he will identify the two faulty machines before going home, and repair them next morning. It takes him twenty minutes to walk to the bus stop. The last bus leaves at 8:32 pm. If it takes six minutes to identify whether a machine is defective or not, and if he decides to check the machines at random, what is the probability that the mechanic will be able to catch the last bus?

a) 0

b) 1/6

c) 1/4

d) 1/3

e) 1

Question 6: 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 7: 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 8: 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 9: 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 10: 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) $\geq1.25$

Let $p \times \frac{0.34 mt}{mt}$m packets of milk be prepared in unit time at the normal speed.

Now, at normal speed in $t$ time, the number of packets of milk that would be produced = $mt$

=> Number of packets of milk produced at fast speed = $(\frac{110}{100} \times m) + (\frac{60}{100} \times t) = 0.66 mt$

The target for the supervisor = $mt$ packets

Number of packets produced at normal speed = $mt – 0.66 mt = 0.34 mt$

Let the probability of a packet being damaged when produced at normal speed = $p$

=> Probability that a packet is damaged when produced at fast speed = $2p$

The probability that a packet selected at random will be damaged = 0.112

=> $(p \times \frac{0.34 mt}{mt}) + (2p \times \frac{0.66 mt}{mt}) = 0.112$

=> $0.34p + 1.32p = 1.66 p = 0.112$

=> $p = \frac{0.112}{1.66} = 0.067$

$\therefore$ Probability that a packet will not be damaged at normal speed = $1 – 0.067 = 0.93$

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 = $\frac{3}{8} \times \frac{1}{2}$

= $\frac{3}{16}$

As the mechanic has decided to check two machines, thus if he identifies either both defective machines or  both non defective, then the probability that he is able to catch the last bus is sum of the two. Thus, the possible outcomes are :

(i) : Both are defective machines, probability = $(\frac{1}{2})(\frac{1}{3}) = \frac{1}{6}$

(ii) : First is defective and second is non defective, probability = $(\frac{1}{2})(\frac{2}{3}) = \frac{1}{3}$

(iii) : First is non defective and second is defective, probability = $(\frac{1}{2})(\frac{2}{3}) = \frac{1}{3}$

(iv) : Both are non defective machines, probability = $(\frac{1}{2})(\frac{1}{3}) = \frac{1}{6}$

In the first and last cases, the mechanic would have identified the defective machines in time to catch the bus.

$\therefore$ Probability that he is able to catch the last bus = $\frac{1}{6} + \frac{1}{6} = \frac{1}{3}$

=> Ans – (D)

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.

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.

We hope this Probability Questions PDF for CAT with Solutions will be helpful to you.