The probability mass function of random variable $\mathrm{X}$ is given by…

The probability mass function of random variable $\mathrm{X}$ is given by $\mathrm{P}[\mathrm{X}=\mathrm{r}]=\left\{\begin{array}{ll} \frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}}{32}, & \mathrm{n}, \mathrm{r} \in \mathbb{N} \\ 0, & \text { otherwise } \end{array} \text {, then } \mathrm{P}[\mathrm{X} \leq 2]=\right.$
  1. $\frac {1}{3}$
  2. $\frac {1}{2}$
  3. $\frac {1}{4}$
  4. $\frac {1}{5}$

Solution

$\begin{aligned} & \text { Since } \sum_{x=0}^{\mathrm{n}} \mathrm{P}(\mathrm{X}=\mathrm{n})=1 \\ & \frac{{ }^n C_0+{ }^n C_1+{ }^n C_2 \ldots+\ldots+{ }^n C_n}{32}=1 \\ & 2^{\mathrm{n}}=32 \\ & \therefore \quad \mathrm{n}=5 \\ & \mathrm{P}(\mathrm{X} \leq 2)=\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2) \\ & =\frac{{ }^5 \mathrm{C}_0}{32}+\frac{{ }^5 \mathrm{C}_1}{32}+\frac{{ }^5 \mathrm{C}_2}{32}=\frac{1}{2} \\ & \end{aligned}$

Asked in: MHT CET 2023 (09 May Shift 1)

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