Question

Suppose a box contains 3 defective transistors and 12 good transistors.if two transistors are drawn from the box without replacement what is the probability that.

(I)the first is good and second transistor is defective?

(II)the first transistor is defective and the second transistor is good?

(III)one of the drawn transistor is good and the other is defective?.

Answer 1

$\text{the probability that:}\\ \text{(I)the first is good and second }\\ \text{transistor is defective}\\ =\frac{12}{15} \times \frac{3}{14}=\frac{36}{210}\\ \text{(II)the first transistor is defective and}\\ \text{ the second transistor is good}\\ =\frac{3}{15} \times \frac{12}{14}=\frac{36}{210}\\ \text{(III)one of the drawn transistor}\\ \text{ is good and the other is defective}\\ =\frac{12}{15} \times \frac{3}{14}+\frac{3}{15} \times \frac{12}{14}\\ =\frac{72}{210}$