# Probability and Statistics (Test 6)

## Cat Entrance Exams : Mathematics Or Quantitative Aptitude

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Q.1
Direction: Study the given information carefully and answer the questions that follow—
A store contains 4 red, 5 blue, 4 green shirts.
If two shirts are picked at random, what is the probability that both are green?
A. 3/19
B. 1/17
C. 1/13
D. 1/11
E. None of these
Explaination / Solution:

Probabilities if both are green
4C2/13C2 = 6/78 → 1/13

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Q.2

Direction: Study the following information carefully and answer the questions that follow:

A committee of five members is to be formed out of 3 trainees, 4 professors and 6 engineers.

In how many ways this can be done if at least 3 trainees to have to be included in a committee

A. 25
B. 95
C. 65
D. 45
E. None of these
Explaination / Solution:

Number of Ways when if at least 3 trainees include in committee =

3C3*10C2 = 1*45 --> 45

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Q.3
Directions: Study the given information carefully and answer the questions that follow—
A store contains 5 red, 4 blue, 5 green shirts.
If two shirts are picked at random, what is the probability that both are blue?
A. 3/91
B. 1/17
C. 6/91
D. 1/31
E. None of these
Explaination / Solution:

Probabilities if both are blue
4C2/14C2 = 6/91

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Q.4
In a basket there are 7 apples and 8 oranges. 4 fruits are picked at random. What is the probability that two fruits are apples and 2 are oranges?
A. 72/455
B. 82/455
C. 89/455
D. 84/455
E. None of these
Explaination / Solution:

Total number of fruits n(s) =7 + 8=15
Probability  = (7C2*8C2)/15C4
= (21*28)/1365 = 28/65

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Q.5
If four bats are picked at random, what is the probability that two are blue and two is green?
A. 12/455
B. 35/355
C. 18/455
D. 18/35
E. None of these
Explaination / Solution:

Probability if two are Blue and two are Green =
[(4C2*4C2)/15C4] = (6*6)/1365 → 12/455

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Q.6
Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’.
A. 0.2
B. 1
C. 0
D. 0.5
Explaination / Solution:

S = {(3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6), (1,H), (2,H), (4,H), (5,H), (1,T), (2,T), (4,T), (5,T)} Let A = event that coin shows a tail. i.e. A = { (1,T), (2,T), (4,T), (5,T)} and B = event that atleast one die shows 3. B = {(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),(6,3)} Workspace
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Q.7
The conditional probability of an event E, given the occurrence of the event F lies between
A. 0 < P (E|F) ≤ 1
B. 0 < P (E|F) < 1
C. 0 ≤ P (E|F) ≤ 1
D. 0 ≤ P (E|F) < 1
Explaination / Solution:

As the probability of any event always lies between 0 and 1. Therefore , 0 ≤ P (E|F) ≤ 1.

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Q.8
If P (A) = 0.8, P (B) = 0.5 and P(B|A) = 0.4, find P(A ∪ B)
A. 0.98
B. 1.00
C. 0.95
D. 0.25
Explaination / Solution: Workspace
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Q.9
A pair of dice is tossed once and a total of 8 has come up. The chance that both the dice show up same number is
A. 5/ 216
B. none of these
C. 1/ 5
D. 1/ 6
Explaination / Solution:

on tossing a pair of  dice total outcomes are 36

out of which getting a total of 8 have possiblities {(2,6),(6,2),(3,5),(5,3),(4,4)}=5

and from these 5 outcomes getting same no. on both dice is (4,4)=1

so, probability is 1/5

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Q.10
8 coins are tossed at a time. The probability of getting atleast 6 heads up is
A. 1/ 64
B. 229/ 256
C. 57/ 64
D. 37/ 256