Question

A box contains 10 balls, of which 3 are red, 2 are yellow, and 5 are blue. Five balls are randomly selected with replacement. Calculate the probability that fewer than 2 of the selected balls are red?

Options

• 0.3601

• 0.5000

• 0.5282

• 0.8369

Answer 1

0.5282

Explanation:

Total number of balls = 10

Red = 3, Blue = 5 and Yellow = 2

Probability of getting red ball, p = `3/10`

Probability of not getting red ball, q = `7/10`

Five balls are randomly selected, n = 5

⇒ Now, probability that fewer than 2 of the selected balls are red, P(X = 0) + P(X = 1)

= `""^5C_0 (3/10)^0 (7/10)^5 + ""^5C_1 (3/10)^1 (7/10)^4`

= `7^5/10^5 + (3 xx 7^4 xx 5)/10^5`

= `(7^4 (7 + 15))/10^5`

= `(22 xx 74)/10^5`

∴ Probability = 0.5282.