In a box of 100 bulbs 30 are defective. What is the probability of getting more than 2 defective bulbs if 5 are chosen randomly from the box?

Question

Question

This question was previously asked in

SSC CGL Tier-II ( JSO ) 2017 Official Paper ( Held On : 22 Feb 2018 )

Answer 1

Calculation

Total number of bulbs = 100

Defective bulb = 30

Probability of defective bulb = p = 30/100

⇒ p = 3/10

⇒ q = 1 – p

⇒ q = 1 – 3/10 = 7/10

When 5 bulb are chosen random, n = 5

P(X = x) = ⁿc_xp^xq^{n – x}

⇒ ⁵c_xp^xq^{5 – x}

Getting more than 2 defective bulbs, x = 3, 4, 5

⇒ ⁵c₃(3/10)³(7/10)² + ⁵c₄(3/10)⁴(7/10) + ⁵c₅(3/10)⁵(7/10)^{5 – 5}

⇒ 0.1323 + (5 × 81 × 7)/100000 + (243/100000)

⇒ 0.1323 + 0.02835 + 0.00243

⇒ 0.163

∴ Probability of getting more than 2 defective bulbs if 5 are chosen randomly from the box is 0.163