**Probability Questions for SNAP**

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**Question 1: **The supervisor of a packaging unit of a milk plant is being pressurized 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: **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: **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 5: **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 6: **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

**Question 7: **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 8: **In a reputed engineering college in Delhi, students are evaluated based on trimesters. The probability that an Engineering student fails in the first trimester is 0.08. If he does not fail in the first trimester, the probability that he is promoted to the second year is 0.87. The probability that the student will complete the first year in the Engineering College is approximately:

a) 0.8

b) 0.6

c) 0.4

d) 0.7

**Instructions**

Answer the questions based on the following information.

Rajat is the sales manager of Dubin Computers Ltd. and looks after the Delhi market. The company sells laptops in India. He is currently trying to select a distributor for the coming five years. The distributor ensures that the products are accessible to the customers in the market. Market share of a company depends on the coverage by the distributor. The total profit potential of the entire laptop market in Delhi is Rs. 5 crores in the current year and the present value of the next four years’ cumulative profit potential is Rs. 15 crores. The first choice for Rajat is to enter into a long-term contract with a distributor M/s Jagan with whom Dubin has done business in the past, and whose distribution system reaches 55 percent of all potential customers. At the last moment, however, a colleague suggests Rajat to consider signing a one-year contract with other distributors. Distributors M/s Bola and M/s James are willing to be partner with Dubin. Although a year ago M/s Bola’s and M/s James’s coverage reached only 40 and 25 percent of customers respectively, they claim to have invested heavily in distribution resources and now expect to be able to reach 60 percent and 75 percent of customers respectively. The probability of M/s Bola’s claim and M/s James’s claim to be true is 0.60 and 0.20 respectively. The knowledge about distributors’ coverage will evolve over time. The assumption is that the true level of coverage offered by the new distributors could be discovered, with certainty, through a one-year trial, and this trail will reveal exactly one of the two levels of coverage: for example in case of M/s Bola – 40 percent (as it was last year) or 60 percent (as claimed). In addition, it is also assumed that whatever the coverage is for both distributors, it will not change over time. Rajat narrows down on three choices, which are as follows:

Choice 1. Give a five-year contract to the familiar distributor M/s Jagan.

Choice 2. Give a one year contract to the new distributor M/s Bola, and base next year’s decision to renew contract with M/s Bola on observed coverage for the next four years or enter into a four years’ contract with M/s Jagan.

Choice 3. Give a one-year contract to the new distributor M/s James, and base next year’s decision to renew the contract with M/s James on observed coverage for next four years or enter into a four years contract with M/s Jagan.

**Question 9: **If the distributor M/s James claims a coverage of 55% instead of 75% and probability of this claim to be true is 0.70 instead of 0.20 then which of the following statement is true?

a) Choice 1 is more profitable than Choice 2

b) Choice 2 is more profitable than Choice 3

c) Choice 3 is more profitable than Choice 1

d) None of the above

**Question 10: **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

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**Answers & Solutions:**

**1) Answer (B)**

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 = $1 – 0.067 = 0.93$

**2) Answer (D)**

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

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

$\therefore b – a = 30 – 7 = 23$

**3) Answer (C)**

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

= $\frac{3}{16}$

**4) Answer (A)**

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.

**5) Answer (C)**

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.

**6) Answer (E)**

The radius of the coin is 3 cm.

So, if the coin should not cross the edge, the centre of the coin should at least be 3 cm away from the edge of the tile.

In the given diagram the center of the circle should lie in the blue region. As we can see, the area in which the centre of the coin can fall is a square of side 10-3-3 = 4 cm.

Therefore, the area in which the centre of the coin can fall is 16 square cm.

Area of the tile = 100 square cm.

Required probability = 16/100 = 0.16.

Therefore, option E is the right answer.

**7) Answer (E)**

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.

**8) Answer (A)**

The probability that the student passes in the first trimester is 0.92

Now given that if the student passes in the first trimester then probability of moving into second second year is 0.87

Hence, the probability of completing first year is 0.92 x 0.87 = 0.80

**9) Answer (B)**

We are left with 3 choices.

Choice 1:

The first choice is to give the contract to M/S Jagan. In this case, we know that Jagan’s market reach is 55%. It has been given that the total profit potential is 5 crores in the present year and 15 crores in the next 4 years.

Therefore, the expected value of profit earned for choice 1 is 0.55*(5+15) = Rs.11 crore.

Choice 2:

Give the contract to M/s Bola for one year and based on the performance, renew the contract with him for the next 4 years or give M/S Jagan the contract for the next 4 years.

Let us assume that M/S Bola retains the contract for all 5 years. Rajat will renew the contract only if M/S Bola’s claim that their market reach is 60% is true. The probability of the claim being true is 0.6.

Therefore, the EV of return if M/S Bola bags the contract for all 5 years = 0.6*0.6*(5+15) = Rs. 7.2 crores.

Let us assume that M/S Bola’s claim is false. The probability of the claim being false is 1-0.6 = 0.4.

Now, if the claim is false, Rajat will terminate the contract by the end of the year and will partner with M/S Jagan for the next 4 years. Also, we have historic data that M/S Bola reaches 40% of the customers. Even if the claim is false, the laptops will reach 40% of the customers in the first year and 55% of the customers from the second year (Since M/S Jagan will bag the contract).

Therefore, the EV of profit in this case is 0.4*0.4*5+0.4*0.55*15 = 0.8 + 3.3 = Rs.4.1 crores.

Therefore, the total EV if M/S Bola bags the contract the first year is 7.2+4.1 = Rs.11.3 crores.

**It has been given in this question that M/S James claims a coverage of 55% and the probability of this being true is 0.7.**

Choice 3:

Give the contract to M/s James for one year and based on the performance, renew the contract with him for the next 4 years or give M/S Jagan the contract for the next 4 years.

Let us assume that M/S James retains the contract for all 5 years. Rajat will renew the contract only if M/S Jame’s claim that their market reach is 55% is true. The probability of the claim being true is 0.7.

Therefore, the EV of return if M/S James bags the contract for all 5 years = 0.7*0.55*(5+15) = Rs. 7.7 crores.

Let us assume that M/S James’s claim is false. The probability of the claim being false is 1-0.7 = 0.3.

Now, if the claim is false, Rajat will terminate the contract by the end of the year and will partner with M/S Jagan for the next 4 years. Also, we have historic data that M/S James reaches 25% of the customers. Even if the claim is false, the laptops will reach 25% of the customers in the first year and 55% of the customers from the second year (Since M/S Jagan will bag the contract).

Therefore, the EV of profit in this case is 0.3*0.25*5+0.3*0.55*15 = 0.375 + 2.475 = Rs.2.85 crores.

Therefore, the total EV if M/S Bola bags the contract the first year is 7.7+2.85 = Rs.10.55 crores.

EV of choice 1 = Rs. 11 crores

EV of choice 2 = Rs. 11.3 crores

EV of choice 3 = Rs. 10.55 crores

Arranging the choices by EV, we get, Choice 2 > Choice 1 > Choice 3.

Choice 2 is more profitable than choice 3. Therefore, option B is true and hence, option B is the right answer.

**10) Answer (B)**

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.

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