Q99 / 150 · MHT CET 2026 (13 April Shift 2) · Mathematics
It is known that a box of $8$ batteries contains $3$ defective pieces and a person randomly selects $2$ batteries from this box. Then the probability distribution of the number of defective batteries is
- Correct answerA
$X = x$ $0$ $1$ $2$ $P(X = x)$ $\dfrac{10}{28}$ $\dfrac{15}{28}$ $\dfrac{3}{28}$ - B
$X = x$ $1$ $2$ $3$ $P(X = x)$ $\dfrac{10}{28}$ $\dfrac{15}{28}$ $\dfrac{3}{28}$ - C
$X = x$ $0$ $1$ $2$ $P(X = x)$ $\dfrac{15}{28}$ $\dfrac{10}{28}$ $\dfrac{3}{28}$ - D
$X = x$ $1$ $2$ $3$ $P(X = x)$ $\dfrac{15}{28}$ $\dfrac{10}{28}$ $\dfrac{3}{28}$
Solution
Total number of batteries = $8$
Number of defective batteries = $3$
Number of non-defective batteries = $5$
Let $X$ be the random variable representing the number of defective batteries selected. Since $2$ batteries are drawn, the possible values of $X$ are $0, 1, 2$.
Total number of ways to select $2$ batteries from $8$ is $^{8}C_{2} = 28$.
Probability of selecting $0$ defective batteries ($X = 0$):
$P(X = 0) = \dfrac{^{5}C_{2}}{^{8}C_{2}} = \dfrac{10}{28}$
Probability of selecting $1$ defective battery ($X = 1$):
$P(X = 1) = \dfrac{^{3}C_{1} \times ^{5}C_{1}}{^{8}C_{2}} = \dfrac{3 \times 5}{28} = \dfrac{15}{28}$
Probability of selecting $2$ defective batteries ($X = 2$):
$P(X = 2) = \dfrac{^{3}C_{2}}{^{8}C_{2}} = \dfrac{3}{28}$
Thus, the probability distribution is:
$X = 0, P(X = 0) = \dfrac{10}{28}$
$X = 1, P(X = 1) = \dfrac{15}{28}$
$X = 2, P(X = 2) = \dfrac{3}{28}$
Answer:
Number of defective batteries = $3$
Number of non-defective batteries = $5$
Let $X$ be the random variable representing the number of defective batteries selected. Since $2$ batteries are drawn, the possible values of $X$ are $0, 1, 2$.
Total number of ways to select $2$ batteries from $8$ is $^{8}C_{2} = 28$.
Probability of selecting $0$ defective batteries ($X = 0$):
$P(X = 0) = \dfrac{^{5}C_{2}}{^{8}C_{2}} = \dfrac{10}{28}$
Probability of selecting $1$ defective battery ($X = 1$):
$P(X = 1) = \dfrac{^{3}C_{1} \times ^{5}C_{1}}{^{8}C_{2}} = \dfrac{3 \times 5}{28} = \dfrac{15}{28}$
Probability of selecting $2$ defective batteries ($X = 2$):
$P(X = 2) = \dfrac{^{3}C_{2}}{^{8}C_{2}} = \dfrac{3}{28}$
Thus, the probability distribution is:
$X = 0, P(X = 0) = \dfrac{10}{28}$
$X = 1, P(X = 1) = \dfrac{15}{28}$
$X = 2, P(X = 2) = \dfrac{3}{28}$
Answer:
| $X = x$ | $0$ | $1$ | $2$ |
| $P(X = x)$ | $\dfrac{10}{28}$ | $\dfrac{15}{28}$ | $\dfrac{3}{28}$ |
Hence, the correct answer is option: 'A'.