Algebra

Probability

Medium

applied

There are 50 telephone lines in an exchange. The probability that any one of them will be busy is 0.1, then the probability that all the lines are busy?

$\frac{e^{5} \times 5^{50}}{5!}$

$\frac{e^{-5} \times 5^{50}}{50!}$

$\frac{e^{-50} \times 50^{50}}{50!}$

$\frac{e^{-50} \times 50^5}{50!}$

✅ Correct Option: 2

There are 50 telephone lines, and each line has a probability of 0.1 of being busy. The probability that all 50 lines are busy at the same time needs to be determined.

This is a binomial probability problem where each line can be either busy or not busy, with each line operating independently and the probability remaining constant at 0.1.

Given:

Number of lines (n) = 50
Probability of one line being busy (p) = 0.1
All lines busy, so k = 50

When dealing with a large number of trials (50 lines) and a small probability (0.1), the Poisson approximation can be applied.

Calculate $\lambda$ :

$\lambda = n \times p$

$\lambda = 50 \times 0.1$

$\lambda = 5$

The Poisson probability formula is:

$P(X = k) = \frac{e^{-\lambda} \times \lambda^k}{k!}$

For $\lambda = 5$ and $k = 50$ :

$P(X = 50) = \frac{e^{-5} \times 5^{50}}{50!}$

The probability that all lines are busy is $\frac{e^{-5} \times 5^{50}}{50!}$

2026: 15 May Shift 1

Three balls are drawn one by one without replacement from a bag containing 5 white and 4 red balls. Let X denote the number of white balls.

Match the LIST-I with LIST-II

LIST-I	LIST-II
A. $P(X=0)$	I. $10/21$
B. $P(X=1)$	II. $5/42$
C. $P(X=2)$	III. $1/21$
D. $P(X=3)$	IV. $5/14$

Choose the correct answer from the options given below:

2022: 7 Aug Shift 2

A shopkeeper sells three types of flower seeds $A_1, A_2, A_3$ . They are sold as a mixture where the proportions are 4 : 4 : 2 respectively. The germination rates of the three types of seeds are 45%, 60% and 35% respectively. Calculate the probability in the following cases.

The probability that seed is not of type $A_1$ , given that seed germinates.

2025: 26 May Shift 2

The random variable X has the following probability distribution

X	0	1	2	3
P(X)	a	a	b	b

such that E(x²) = 2E(x), then the value of b is:

2026: 14 May Shift 2

Four cards are drawn successively without replacement from a deck of 52 playing cards. The probability that all the four cards are king is:

2025: 30 May Shift 1

A random variable y has the following probability distribution

y	1	2	3	4	5
P(y)	2k	3k	k	4k	5k

Match List-I with List-II

List-I	List-II
(A) $P(y > 2)$	(I) $2/5$
(B) k	(II) $2/3$
(C) $P(y \leq 3)$	(III) $8/15$
(D) $P(2 \leq y \leq 4)$	(IV) $1/15$

Choose the correct answer from the options given below:

2025: 19 May Shift 1

Match List-I with List-II

Let A and B be any two events

List-I	List-II
(A) $P(A')$	(I) $\frac{P(A \cap B)}{P(A)}; P(A) \neq 0$
(B) $P(\phi)$	(II) $\frac{P(A \cap B)}{P(B)}; P(B) \neq 0$
(C) $P(A\vert B)$	(III) $1 - P(A)$
(D) $P(B\vert A)$	(IV) 0

Choose the correct answer from the options given below:

There are 50 telephone lines in an exchange. The probability that any one of them will be busy is 0.1, then the probability that all the lines are busy?

Solution

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