MHTCET 13 Apr 2026 Shift 2 Paper

Concept:Probability distribution of a discrete random variable using combinations from a finite set.Explanation:Total batteries in the box $= 8$.Defective batteries $= 3$, non-defective batteries $= 8 - 3 = 5$.Let $X$ be the number of defective batteries selected when 2 batteries are drawn randomly.Possible values of $X$ are $0, 1, 2$.Total ways to select 2 batteries from 8:$${}^{8}C_{2} = \frac{8 \times 7}{2} = 28$$For $X = 0$: both selected batteries are non-defective.$$P(X=0) = \frac{{}^{5}C_{2}}{{}^{8}C_{2}} = \frac{10}{28}$$For $X = 1$: one defective and one non-defective battery are selected.$$P(X=1) = \frac{{}^{3}C_{1} \times {}^{5}C_{1}}{{}^{8}C_{2}} = \frac{15}{28}$$For $X = 2$: both selected batteries are defective.$$P(X=2) = \frac{{}^{3}C_{2}}{{}^{8}C_{2}} = \frac{3}{28}$$Hence the distribution for $X = 0, 1, 2$ has probabilities $\frac{10}{28}, \frac{15}{28}, \frac{3}{28}$ respectively.Answer:The correct probability distribution is given in Option A.