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The probability distribution of random variable $X$ with two missing probabilities $p_{1}$ and $p_{2}$ is given below
$\begin{array}{cc}\boldsymbol{X} & \boldsymbol{P}(\boldsymbol{X}) \\ 1 & k \\ 2 & p_{1} \\ 3 & 4 k \\ 4 & p_{2} \\ 5 & 2 k\end{array}$
It is further given that $P(X \leq 2)=0.25$ and $P(X \geq 4)=0.35$. Consider the following statements
$1.$ $p_{1}=p_{2}$
$2.$ $p_{1}+p_{2}=P(X=3)$
Which of the statements given above is/are correct?
Options:
$\begin{array}{cc}\boldsymbol{X} & \boldsymbol{P}(\boldsymbol{X}) \\ 1 & k \\ 2 & p_{1} \\ 3 & 4 k \\ 4 & p_{2} \\ 5 & 2 k\end{array}$
It is further given that $P(X \leq 2)=0.25$ and $P(X \geq 4)=0.35$. Consider the following statements
$1.$ $p_{1}=p_{2}$
$2.$ $p_{1}+p_{2}=P(X=3)$
Which of the statements given above is/are correct?
Solution:
2940 Upvotes
Verified Answer
The correct answer is:
Neither 1 nor 2
Let $P(X \leq 2)=0.25$
$\Rightarrow \quad P(X=1)+P(X=2)=0.25$
$\Rightarrow k+p_{1}=0.25($ from the table $)$
$\Rightarrow p_{1}=0.25-k$
and $P(X \geq 4)=0.35$
$\Rightarrow \quad P(X=4)+P(X=5)=0.35$
$\Rightarrow p_{2}+2 k=0.35$ (from the table)
$\Rightarrow p_{2}=0.35-2 k$
From (1) and (2)
$p_{1} \neq p_{2}$
and $p_{1}+p_{2}=0.25-k+0.35-2 k=0.6-3 k$
$\neq P(X=3)$
Hence, neither 1 nor 2 is correct
$\Rightarrow \quad P(X=1)+P(X=2)=0.25$
$\Rightarrow k+p_{1}=0.25($ from the table $)$
$\Rightarrow p_{1}=0.25-k$
and $P(X \geq 4)=0.35$
$\Rightarrow \quad P(X=4)+P(X=5)=0.35$
$\Rightarrow p_{2}+2 k=0.35$ (from the table)
$\Rightarrow p_{2}=0.35-2 k$
From (1) and (2)
$p_{1} \neq p_{2}$
and $p_{1}+p_{2}=0.25-k+0.35-2 k=0.6-3 k$
$\neq P(X=3)$
Hence, neither 1 nor 2 is correct
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