Number of defective bulbs in a lot of 500 bulbs follows a binomial distribution with probability of a randomly selected bulb to be defective equal to 0.3 A. A sample of 50 bulbs is drawn. Probability of 2 defective bulbs in the sample is :

$1225 (0.3)^2 (0.7)^{48}$

The correct answer is Option (3) → $1225 (0.3)^2 (0.7)^{48}$

No. of drawn = 50

S = success = defective bulb drawn

$P(S) = 0.3$

$P(\overline S)=0.7$

$P(S=2)={^{50}C}_2×0.3^2×0.7^{48}$

$=\frac{50×49}{2}×0.3^2×0.7^{48}$

$=1125(0.3)^2(0.7)^{48}$