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Probability

Multiple Choice Questions

91.

Seven white balls and three black balls are randomly arranged in a row. The probability that no two black balls are placed adjacently is

$\frac{1}{2}$
$\frac{7}{15}$
$\frac{2}{15}$
$\frac{1}{3}$

92.

The probability distribution of a random variable X is given below

X = x	0	1	2	3
P(X) = x	$\frac{1}{10}$	$\frac{2}{10}$	$\frac{3}{10}$	$\frac{4}{10}$

Then the variance of X is

93.

The probability that an individual suffers a bad reaction from an injection is 0.001. The probability that out of 2000 individuals exactly three will suffer bad reaction is

$\frac{1}{e^{2}}$
$\frac{2}{3 e^{2}}$
$\frac{8}{3 e^{2}}$
$\frac{4}{3 e^{2}}$

94.

A regular polygon of n sides has 170 diagonals,then n is equal to

95.

A committee of 12 members is to be formed from 9 women and 8 men. The number of committees in which the women are in majority is

2720
2702
2270
2278

96.

A student has to answer 10 out of 13 questions in an examination choosing atleast 5 questions from the first 6 questions. The number of choice available to the student is

97.
In an entrance test there are multiple choice questions. There are four possible answers to each question, of which one is correct. The probability that a student know the answer to a question is 9/10. If he gets the correct answer to a question, then the probability that he was guessing is

$\frac{37}{40}$

$\frac{1}{37}$

$\frac{36}{37}$

$\frac{1}{9}$

$\frac{1}{37}$

$Let E_{1} : The event that the students knows the answer and E_{2} : The event that the student guesses the answer ∴ P (E_{1}) = \frac{9}{10} and P (E_{1}) = 1 - \frac{9}{10} Let E : The answer is correct . The probability that the student answered correctly, given that he knows the answer i . e . P (\frac{E}{E_{1}}) = 1 The probability that the students answered correctly, given that he guessed is \frac{1}{4}$

$P (\frac{E}{E_{2}}) = \frac{1}{4} By using Baye' s theorem, P (\frac{E}{E_{2}}) = \frac{P (\frac{E}{E_{2}}) P (E_{2})}{P (\frac{E}{E_{1}}) P (E_{1}) + P (\frac{E}{E_{2}}) P (E_{2})} = \frac{\frac{1}{4} \times \frac{1}{10}}{1 \times \frac{9}{10} + \frac{1}{4} \times \frac{1}{10}} = \frac{\frac{1}{40}}{\frac{9}{10} + \frac{1}{40}} = \frac{\frac{1}{40}}{\frac{36 + 1}{40}} = \frac{1}{37}$

98.

There are four machines and it is known that exactly two of them are faulty. They are teste done by one, in a random order till both the faulty machines are identified. Then, the probability that only two tests are need is

$\frac{1}{3}$
$\frac{1}{6}$
$\frac{1}{2}$
$\frac{1}{4}$

99.

A random variable X has the probability distribution given below.

X	1	2	3	4	5
P(X = x)	K	2K	3K	2K	K

Its variance is

$\frac{16}{3}$
$\frac{4}{3}$
$\frac{5}{3}$
$\frac{10}{3}$

100.

A candidate takes three tests in succession and the probability of passing the first test is p. The probability of passing each succeeding test is p or $\frac{p}{2}$ according as he passes or fails in the preceding one. The candidate is selected, if he passes atleast two tests. The probability that the candidate is selected, is

p²(2 - p)
p(2 - p)
p + p² + p³
p²(1 - p)