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# Probability Questions for NMAT:

Download Probability Questions for NMAT PDF. Top 10 very important Probability Questions for NMAT based on asked questions in previous exam papers.

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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: 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: The scheduling officer for a local police department is trying to schedule additional patrol units in each of two neighbourhoods – southern and northern. She knows that on any given day, the probabilities of major crimes and minor crimes being committed in the northern neighbourhood were 0.418 and 0.612, respectively, and that the corresponding probabilities in the southern neighbourhood were 0.355 and 0.520. Assuming that all crime occur independent of each other and likewise that crime in the two neighbourhoods are independent of each other, what is the probability that no crime of either type is committed in either neighbourhood on any given day?

a) 0.069

b) 0.225

c) 0.69

d) 0.775

e) None of the above

Question 4: A doctor has decided to prescribe two new drugs D1and D2 to 200 heart patients such that 50 get drug D1, 50 get drug D2 and 100 get both. The 200 patients are chosen so that each had 80% chance of having a heart attack if given neither of the drugs. Drug D1 reduces the probability of a heart attack by 35 %, while drug D2 reduces the probability by 20%. The two drugs when taken together, work independently. If a patient, selected randomly from the chosen 200 patients, has a heart attack then the probability that the selected patient was given both the drug is:

a) 0.42

b) 0.49

c) 0.56

d) 0.40

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: A survey was conducted to test relative aptitudes in quantitative and logical reasoning of MBA applicants. It is perceived (prior to the survey) that 80 percent of MBA applicants are extremely good in logical reasoning, while the other 20 percent are extremely good in quantitative aptitude. Further, it is believed that those with strong quantitative knowledge are also sound in data interpretation, with conditional probability as high as 0.87. However, some MBA applicants who are extremely good in logical reasoning can be also good in data interpretation, with conditional probability 0.15. An applicant surveyed is found to be strong in data interpretation. The probability that the applicant is also strong in quantitative aptitude is

a) 0.4

b) 0.6

c) 0.8

d) 0.9

Question 7: The game of “chuck-a-luck” is played at carnivals in some parts of Europe. Its rules are as follows: if you pick a number from 1 to 6 and the operator rolls three dice. If the number you picked comes up on all three dice, the operator pays you €3; if it comes up on two dice, you are paid €2; and if it comes up on just one die, you are paid €1. Only if the number you picked does not come up at all, you pay the operator €1. The probability that you will win money playing in this game is:

a) 0.52

b) 0.753

c) 0.42

d) None of these

Question 8: McDonald’s ran a campaign in which it gave game cards to its customers. These game cards made it possible for customers to win hamburgers, French fries, soft drinks, and other fast-food items, as well as cash prizes. Each card had 10 covered spots that could be uncovered by rubbing them with a coin. Beneath three of these spots were “No Prize” signs. Beneath the other seven spots were names of the prizes, two of which were identical. For, example, one card might have two pictures of a hamburger, one picture of a coke, one of French fries, one of a milk shake, one of a $5, one of$1000, and three “No Prize” signs. For this card the customer could win a hamburger. To win on any card, the customer had to uncover the two matching spots (which showed the potential prize for that card)before uncovering a “No Prize”; any card with a “No Prize” uncovered was automatically void. Assuming that the two matches and the three “No Prize” signs were arranged randomly on the cards, what is the probability of a customer winning?

a) 0.10

b) 0.15

c) 0.12

d) 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: The probability that a leap year selected at random contains either 53 Sundays or 53 Mondays, is:

a) $\frac{17}{53}$

b) $\frac{1}{53}$

c) $\frac{3}{7}$

d) None of these

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 = $1 – 0.067 = 0.93$

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

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

$\therefore b – a = 30 – 7 = 23$

For northern neighbourhood,

Probability that there is no major crime = $(1 – 0.418) = 0.582$

Probability that there is no minor crime = $(1 – 0.612) = 0.388$

For southern neighbourhood,

Probability that there is no major crime = $(1 – 0.355) = 0.645$

Probability that there is no minor crime = $(1 – 0.520) = 0.480$

$\therefore$ Probability that no crime of either type is committed in either neighbourhood on any given day

= $0.582 \times 0.388 \times 0.645 \times 0.480$

= $0.069$

Given that probability of getting a heart attack before any drug = 0.80

Drug D1 reduces the probability of a heart attack by 35 %. Therefore, the probability of a patient getting heart attack after he has taken D1 = (1-0.35)*0.80 = 0.52

Drug D2 reduces the probability of a heart attack by 20 %. Therefore, the probability of a patient getting heart attack after he has taken D2 = (1-0.20)*0.80 = 0.64

It is given that both D1 and D2 work independently. Therefore, the probability of a patient getting heart attack after he has taken both D1 and D2 = (1-0.35)*(1-0.20)*0.80 = 0.416

A total of 100 patients have taken both the drugs whereas only 50-50 patients took drug D1 and D2.

Hence, the probability that the selected patient was given both the drug is = $\dfrac{0.416}{0.416+0.5*0.52+0.5*0.64}$ = 0.417 $\approx$ 0.42

Therefore, we can say that option A 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.

We are given that 80% MBA applicants are extremely good in logical reasoning. Also, some MBA applicants who are extremely good in logical reasoning can be also good in data interpretation, with conditional probability 0.15.

Therefore, the probability of that an applicant is sound in data interpretation as well as logical reasoning = 0.80*0.15 = 0.12

We are given that the other 20 percent are extremely good in quantitative aptitude. Also, some MBA applicants who are extremely good in quantitative aptitude can be also good in data interpretation, with conditional probability 0.87.

Therefore, the probability of that an applicant is sound in data interpretation as well as quantitative aptitude = 0.20*0.87 = 0.174

An applicant surveyed is found to be strong in data interpretation then probability that the applicant is also strong in quantitative aptitude is =

$\dfrac{0.174}{0.174+0.12} \approx$ 0.60. Hence, option B is the correct answer.

There are 3 ways to win money in the game.
The number you picked can come up in one dice, 2 dice or 3 dice.

The probability of the number you picked coming in all three dice = (1/6)*(1/6)*(1/6) = 1/216
The probability of the number picked coming on 2 dice = 3C2*(5/6)(1/6)(1/6) = 15/216
The probability of the number picked coming on 1 dice = 3C1*(5/6)(5/6)(1/6) = 75/216

Probability of winning = 1/216 + 15/214 + 75/216  = 91/216 = 0.421.
Therefore, option C is the right answer.

Case 2: When we win by uncovering just 3 spots.

_ _ P _ _ _ _ _ _ _

From the first two uncovered spots 1 will show up P. Out of remaining 7 spots, 3 spots will be filled by No prize. Total number of ways = 2C1*7C3*5!

There are a total of 10 spots out of which 3 are of one type (No Prize), 2 are of one time (The one which will give us prize) and 5 are different. Therefore, total number of combination in which we can uncover these spots = $\dfrac{10!}{2!*3!}$

We win if we win the two same card before any of the No prize spot. We can win by uncovering just 2 spots and a maximum of 7 spots. Let ‘P’ denotes the occurrence of winner card.

Case 1: When we win by uncovering just 2 spots.

P P _ _ _ _ _ _ _ _

Out of remaining 8 spots, 3 spots will be filled by No prize and 5 with different signs. Total number of ways = 8C3*5!

Case 2: When we win by uncovering just 3 spots.

_ _ P _ _ _ _ _ _ _

From the first two uncovered spots 1 will show up P. Out of remaining 7 spots, 3 spots will be filled by No prize. Total number of ways = 2C1*7C3*5!

Case 3: When we win by uncovering just 4 spots.

_ _  _ P _ _ _ _ _ _

From the first three uncovered spots 1 will show up P. Out of remaining 6 spots, 3 spots will be filled by No prize. Total number of ways = 3C1*6C3*5!

Case 4: When we win by uncovering just 5 spots.

_ _ _ _ P _ _ _ _ _

From the first four uncovered spots 1 will show up P. Out of remaining 5 spots, 3 spots will be filled by No prize. Total number of ways = 4C1*5C3*5!

Case 5: When we win by uncovering just 6 spots.

_ _ _ _ _ P _ _ _ _

From the first five uncovered spots 1 will show up P. Out of remaining 4 spots, 3 spots will be filled by No prize. Total number of ways = 5C1*4C3*5!

Case 6: When we win by uncovering just 7 spots.

_ _ _ _ _ _ P _ _ _

From the first six uncovered spots 1 will show up P. Out of remaining 3 spots, 3 spots will be filled by No prize. Total number of ways = 6C1*3C3*5!

Hence, the probability that a customer will win = $\dfrac{8C3*5!+2C1*7C3*5!+3C1*6C3*5!+4C1*5C3*5!+5C1*4C3*5!+6C1*3C3*5!}{\dfrac{10!}{2!*3!}}$

$\Rightarrow$ $\dfrac{3!*2!*5!(56+70+60+40+20+6)}{10!}$

$\Rightarrow$ $\dfrac{1}{10}$. Therefore, option A 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.

Total number of days in a leap year = 366

It will contain 52 weeks and 2days

These two days can be (Sunday, Monday); (Monday, Tuesday); (Tuesday, Wednesday); (Wednesday, Thursday); (Thursday, Friday); (Friday, Saturday); (Saturday, Sunday)

For 53 Sundays, probability = $\ \ \frac{2}{7}$

Similarly for 53 Mondays, probability =$\ \ \frac{2}{7}$

This includes one way where Sunday and Monday occur simultaneously (i.e) Sunday, Monday

Probability for this =$\ \ \frac{1}{7}$

Hence required probability = $\ \ \frac{2}{7}$ +$\ \ \frac{2}{7}$-$\ \ \frac{1}{7}$

=$\ \ \frac{3}{7}$