MHTCET 22 Apr 2025 Shift 1 Paper

Concept:The process stops at the first head or when 4 tails occur consecutively.Each toss is an independent fair coin toss, so the probability of head or tail on any toss is $\frac{1}{2}$.Explanation:Let $X$ be the number of tosses until the process stops.For $X=1$: The first toss must be head.$\mathrm{P}(X=1) = \mathrm{P}(\mathrm{H}) = \frac{1}{2}$For $X=2$: The first toss is tail, and the second toss is head.$\mathrm{P}(X=2) = \mathrm{P}(\mathrm{T})\mathrm{P}(\mathrm{H}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$For $X=3$: The first two tosses are tails, and the third toss is head.$\mathrm{P}(X=3) = \mathrm{P}(\mathrm{T})\mathrm{P}(\mathrm{T})\mathrm{P}(\mathrm{H}) = \left(\frac{1}{2}\right)^3 = \frac{1}{8}$For $X=4$: Two cases are possible.Case 1: First three tosses are tails and the fourth toss is head: $\left(\frac{1}{2}\right)^4 = \frac{1}{16}$Case 2: All four tosses are tails: $\left(\frac{1}{2}\right)^4 = \frac{1}{16}$Thus, $\mathrm{P}(X=4) = \frac{1}{16} + \frac{1}{16} = \frac{1}{8}$Check: $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{8} = 1$Answer:The probability distribution of the number of tosses $X$ is:$\mathrm{P}(X=1) = \frac{1}{2}$, $\mathrm{P}(X=2) = \frac{1}{4}$, $\mathrm{P}(X=3) = \frac{1}{8}$, $\mathrm{P}(X=4) = \frac{1}{8}$.