Introduction
Probability is one of important topic in Quantitative Aptitude Section. In Probability Quiz 9 article candidates can find questions with answer. By solving this questions candidates can improve and maintain, speed, and accuracy in the exams. Probability Quiz 9 questions are very useful for different exams such as IBPS PO, Clerk, SSC CGL, SBI PO, NIACL Assistant, NICL AO, IBPS SO, RRB, Railways, Civil Services etc.
Q1
A bag contains 2 red caps, 4 blue caps, 3 yellow caps and 5 green caps. If three caps are picked at random, what is the probability that none is green?
- A. [latex]\frac {2}{13}[/latex]
B. [latex]\frac {3}{13}[/latex]
C. [latex]\frac {1}{13}[/latex]
D. [latex]\frac {5}{13}[/latex]
Total caps = 14
Probability = [latex]^{5}{C}_{0} \times \frac {^{9}{C}_{3}}{^{14}{C}_{3}} = \frac {3}{13}[/latex]
Q2
A box contains 27 marbles some are blue and others are green. If a marble is drawn at random from the box, the probability that it is blue is 1/3. Then how many number of green marbles in the box?
- A. 10
B. 15
C. 14
D. 18
Blue marble – x
[latex] \frac {^{x}{C}_{1}}{^{27}{C}_{1}} = \frac {1}{3}[/latex]
[latex] \frac {x}{27} = \frac {1}{3} \rightarrow x = \frac {27}{3} = 9[/latex]
No of green marbles = Total – Blue marble = 27 - 9 = 18
Q3
In a bag there are 4 white, 4 red and 2 green balls. Two balls are drawn at random.What is the probability that at least one ball is of red colour?
- A. [latex]\frac {4}{3}[/latex]
B. [latex]\frac {7}{3}[/latex]
C. [latex]\frac {1}{3}[/latex]
D. [latex]\frac {2}{3}[/latex]
Total Balls = 10
Other than red ball = [latex]^{6}{C}_{2}[/latex]
[latex]\frac {^{6}{C}_{2}}{^{10}{C}_{2}} = \frac {1}{3} \rightarrow 1 - \frac {1}{3} = \frac {2}{3}[/latex]
Q4
P and Q are sitting in a ring with 11 other persons. If the arrangement of 11 persons is at random, then the probability that there are exactly 4 persons between them?
- A. [latex]\frac {1}{3}[/latex]
B. [latex]\frac {4}{7}[/latex]
C. [latex]\frac {1}{5}[/latex]
D. [latex]\frac {1}{6}[/latex]
Fix the position of P, then Q can be sit in 12 positions, so total possible outcome = 12
Now, exactly 4 persons are sitting between them. This can be done in two ways as shown in figure, so favourable outcomes = 2
So, probability = [latex] \frac {2}{12} = \frac {1}{6}[/latex]
Q5
A six-digit is to be formed from the given numbers 1, 2, 3, 4, 5 and 6. Find the probability that the number is divisible by 4.
- A. [latex]\frac {3}{17}[/latex]
B. [latex]\frac {4}{15}[/latex]
C. [latex]\frac {4}{19}[/latex]
D. [latex]\frac {4}{17}[/latex]
For a number to be divisible by 4, the last two digit should be divisible by 4.
So possible cases – 12, 16, 24, 32, 36, 52, 56, 64 (last two digits)
So favourable outcomes = 24 + 24 + 24 + 24 + 24 + 24 + 24 + 24 = 192
So p = [latex] \frac {192}{720} = \frac {4}{15} [/latex]
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