Algebra

Probability

Medium

applied

In a box containing 100 bulbs, 10 are defective. The probability that out of a sample of 5 bulbs, none is defective is

$\frac{1}{10}$

$\left(\frac{9}{10}\right)^5$

$\left(\frac{1}{2}\right)^5$

$\frac{9}{10}$

✅ Correct Option: 2

Total bulbs = 100

Defective bulbs = 10

Non-defective bulbs = 100 - 10 = 90

Sample size = 5 bulbs

The probability that one bulb selected is non-defective:

$P(\text{one bulb is non-defective}) = \frac{90}{100}$

$= \frac{9}{10}$

For all 5 bulbs to be non-defective, each of the 5 bulbs selected must be non-defective.

When all events must occur together, the probabilities are multiplied:

$P(\text{all 5 non-defective}) = \frac{9}{10} \times \frac{9}{10} \times \frac{9}{10} \times \frac{9}{10} \times \frac{9}{10}$

$= \left(\frac{9}{10}\right)^5$

Therefore, the probability that none of the 5 bulbs is defective is $\left(\frac{9}{10}\right)^5$ .

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.

2023: 30 May Shift 3

Match List - I with List - II.

List - I	List - II
(A) If A and B are mutually exclusive events, then $P(A \cup B) =$	(I) $\frac{P(A \cap B)}{P(B)}, P(B) \neq 0$
(B) If A and B are independent events, then $P(A \cap B) =$	(II) $\frac{P(A \cap B)}{P(A)}, P(A) \neq 0$
(C) If A and B are two events of a sample space of an experiment, then $P(A/B) =$	(III) $P(A) \cdot P(B)$
(D) If A and B are two events of a sample space of an experiment, then $P(B/A) =$	(IV) $P(A) + P(B)$

Choose the correct answer from the options given below :

2026: 22 May Shift 2

A bag contains 8 red and 7 black balls. Two balls are drawn at random. The probability that both the balls are of the same colour is:

2025: 2 June Shift 1

A pair of dice is thrown until the sum of numbers appeared is a perfect square or a non-perfect square sum appeared five times in succession. If random variable $X$ denotes the number of non perfect square sums appeared, then $P(X > 0)$ is

2026: 29 May Shift 1

If E and F are two events such that $P(E \cup F)=P(E)$ , then

2025: 29 May Shift 2

The probabilities of occurrance of two events A and B are 0.45 and 0.20 respectively. The probability of their simultaneous occurrence is 0.06. The probability that neither A nor B occurs is

In a box containing 100 bulbs, 10 are defective. The probability that out of a sample of 5 bulbs, none is defective is

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