Explore topic-wise InterviewSolutions in Current Affairs.

Correct answer is B.

P(A) = 0.3, P(B) = 0.2 and P(A ∩ B) = 0.1

P(\(\overline{A}\cap B\)) = P(B) – P(A ∩ B) = 0.2 – 0.1 = 0.1

Correct answer is B.

Given:

60% of the students read mathematics, 25% biology and 15% both mathematics and biology

That means,

Let the event A implies students reading mathematics,

Let the event B implies students reading biology,

Then, P(A) = 0.6

P(B) = 0.25

P(A∩B) = 0.15

We, need to find P(A/B) = P(A∩B)/ P(B)

\(\Rightarrow\) \(\frac{0.15}{0.25}=\frac{3}{5}\)

Correct answer is C.

P(A) = probability that A can solve the problem

= 3/5

And P(B) = probability that B can solve the problem = 2/3

P(A ∪ B) = P(A) + P(B), As the events are independent

\(\Rightarrow\) P(A ∩ B) = P(A).P(B)

Thus,

\(\Rightarrow\) ⇒ P(A) + P(B) = 3/5 +2/3 – 2/5 = 13/15

Correct answer is B.

The probability of failure of the first component = 0.2 =P(A)

The probability of failure of second component = 0.3 = P(B)

The probability of failure of third component = 0.5 = P(C)

As the events are independent,

The machine will operate only when all the components work, i.e.,

(1-0.2)(1-0.3)(1-0.5) = P(A’)P(B’)P(C’)

In rest of the cases, it won’t work,

So P(AUBUC) = 1 – P(A’∩B’∩C’) = 1 – (0.8).(0.7).(0.5)

\(\Rightarrow\) 1 – 0.28 = 0.72

Probability of getting an even number is = 3/6 = 1/2

Probability of getting an odd number is = 3/6 = 1/2

Variance = npq

\(\Rightarrow 100\times \frac{1}{2}\times \frac{1}{2}\)

\(\Rightarrow 25\)

If A and B are mutually exclusive events then,

P(A) = 0.4, P(B) = X

And P(A ∪ B) = P(A)+P(B) = 0.5 = 0.4 + P(B)

⇒ P(B) = 0.1

If A and B are mutually exclusive events then,

P(A) = 0.4, P(B) = X

And P(A ∪ B) = P(A)+P(B) = 0.5 = 0.4 + P(B)

⇒ P(B) = 0.1

Variance can not be greater than mean as then, q will be greater than 1, which is not possible.

As, np = 6 and npq = 9

q = 3/2 …(not possible)

Given:

n = 20

q = 0.75

p = ?

q = (1 - p)

= 0.75

p = 1 - 0.75

= 0.25

Mean = np

= 20(0.75)

= 15

Correct answer is (c) \(\frac{9}{13}\)

Correct answer is (b) 5

Let X be a binomial variate with parameters in and p.

Mean = np

Variance = npq

Mean - Variance = np – npq

= np(1 - q)

= np² > 0

So, Mean – Variance > 0

⇒ Mean > Variance

So, mean can never be less than variance.

Correct answer is A.

A die is tossed twice,

The probability of getting a 4, 5 or 6 in the first trial is 3/6 = P(A)

The probability of getting a 1, 2, 3 or 4 in the second trial is 4/6 =P(B)

As the events are independent, the probability of these two events together will be, P(A).P(B) = 1/3.

As 7 coins are tossed simultaneously the total number of outcomes are \(2^7=128.\)

The favourable number of outcomes that a tail appears an odd number of times will be, \(7_{C_1}+7_{C_3}+7_{C_5}+7_{C_7}=64.\)

Thus, the probability

\(=\frac{The\,favourable\,outcomes}{The\,total\,number\,of\,outcomes}\)

\(=\frac{64}{128}\)

\(=\frac{1}{2}\)

Hence, the probability is 1/2 .

As the coin is tossed 5 times the total number of outcomes will be \(2^5=32.\)

And we know that the favourable outcomes of a head appearing even number of times will be,

That either the head appears 0, 2 or 4 times so,

The respective probabilities will be:- \(5_{C_0}\) + \(5_{C_2}\) + \(5_{C_4}\) = 16

Thus, the probability

\(=\frac{The\,favourable\,outcomes}{The\,total\, number\, of\, outcomes}\)

\(\Rightarrow=\frac{16}{32}=\frac{1}{2}\)

Hence, the probability is 1/2.

Correct answer is D.

Given:

n = 4

p(getting atleast one head) = ?

• 0 head : HHHH

• 1 head : HTTT

THTT

TTHT

TTTH

• 2 heads : HHTT

HTHT

HTTH

THHT

THTH

TTHH

• 3 heads : HHHT

HHTH

HTHH

THHH

• 4 heads : HHHH

Total ouctomes = 16

Outcomes containing atleast 1 head = 15

P(getting atleast 1 head) = \(\frac{15}{16}\)