# Probability and Statistics (Test 4)

## Problem Solving And Reasoning : Mathematics Or Quantitative Aptitude

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Q.1
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.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 2 trainees and 3 engineer to have to be included in a committee
A. 32
B. 60
C. 45
D. 90
E. None of these
Explaination / Solution:

If 2 trainees and 3 engineers include in committee
3C2*6C3 = 3*20 --> 60

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Q.3
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 2 trainees to have to be included in a committee
A. 340
B. 820
C. 560
D. 405
E. None of these
Explaination / Solution:

If at least 2 trainees include in committee
3C2*10C3 + 3C3*10C2 = 3*120+1*45
= 405

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Q.4
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 1 trainees and 4 engineers be included in a committee
A. 45
B. 32
C. 60
D. 36
E. None of these
Explaination / Solution:

If 1 trainees and 4 engineers include in committee
3C1*6C4 = 3*15 --> 45

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Q.5
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 3 engineers and 2 professors or 2 trainees and 3 professor be included in a committee
A. 140
B. 160
C. 132
D. 155
E. None of these
Explaination / Solution:

If 3 engineers and 2 professors or 2 trainees and 3 professors include in committee
6C3*4C2 + 3C2*4C3 = 20*6 + 3*4 = 120 + 12 = 132

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Q.6
Direction: Study the given information carefully and answer the questions that follow—
A store contains 3 red, 4 blue, 4 green and 4 whites’ bats.
If two bats are picked at random, what is the probability that both are white?
A. 1/3
B. 3/35
C. 3/5
D. 2/35
E. None of these
Explaination / Solution:

Probabilities if both are White
4C2/15C2 = 6/105 → 2/35

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Q.7
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.8
If three bats are picked at random, what is the probability that at least one is green?
A. 2/5
B. 60/65
C. 44/455
D. 58/91
E. None of these
Explaination / Solution:

Probability if at least one is green
[1-(11C3/15C3)] = [1-(165/455)] → 58/91

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Q.9
If two bats are picked at random, what is the probability that either both are red or both are white?
A. 3/5
B. 3/35
C. 14/25
D. 4/105
E. None of these
Explaination / Solution:

Probabilities if both either are Red or either are White
(3C2 + 4C2)/15C2 = (3 + 6)/105 → 3/35

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Q.10
If two bats are picked at random, what is the probability that none is white?
A. 8/21
B. 11/21
C. 13/55
D. 11/22
E. None of these