Question

A fruit box contains 10 apples, 3 of which were defective. Two apples are picked randomly with replacement. The probability that none of the two apples was defective is approximately
1. 0.49
2. 0.25
3. 0.65
4. 0.75

Answer 1

Correct Answer - Option 1 : 0.49

Concept:

The probability of an event A is calculated as

\(\rm P(A) = \frac {Number\ of\ favourable\ outcomes }{Total\ number\ of\ outcomes}\)

Value of P varies in 0 ≤ P ≤ 1.

Calculation:

Given:

Total 10 apples out of which 3 are defective, means 7 are fresh. Two apples picked randomly with replacement. When 1^st fresh apple picked, the probability

P₁ = \(\rm \frac {Number\ of\ favourable\ outcomes }{Total\ number\ of\ outcomes}\) = \(\frac {7}{10}\)

Box now have 6 fresh apples and 3 defective apples. Now, that picked apple is kept again in the box as replacement is done. Again there are 7 fresh and 3 defective apples in the box. When 2^nd fresh apple picked, the probability

P₂ = \(\rm \frac {Number\ of\ favourable\ outcomes }{Total\ number\ of\ outcomes}\) = \(\frac {7}{10}\)

Overall probability:

P = P₁ × P₂ = \(\frac {7}{10} × \frac {7}{10} \) = \(\frac {49}{100}\) = 0.49