If the mean of the following probability distribution of a radam variable X : .tg border-collapse:collapse;border-spacing:0; .tg tdborder-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal; .tg thborder-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal; .tg .tg-baqhtext-align:center;vertical-align:top X 0 2 4 6 8 P(X) a 2a a+b 2b 3b is \frac469 , then the variance of the distribution is

Probability PYQ — JEE Main 2024 · 4 Apr · Shift 2 · Q42 with Solution

If the mean of the following probability distribution of a radam variable $\mathrm{X}$ :

$\mathrm{X}$	0	2	4	6	8
$\mathrm{P(X)}$	$a$	$2a$	$a+b$	$2b$	$3b$

is $\frac{46}{9}$ , then the variance of the distribution is

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Correct answer: B

Given the distribution:

X & 0 & 2 & 4 & 6 & 8 \\ \hline P(X) & a & 2a & a+b & 2b & 3b \end{array}$$ Since total probability is $1$, $$a+2a+(a+b)+2b+3b=1$$ $$4a+6b=1 \quad \Rightarrow \quad 2a+3b=\frac12 \quad ...(1)$$ 2. **Use the mean** Mean is given as $$E(X)=\frac{46}{9}$$ Now, $$E(X)=0\cdot a+2\cdot 2a+4(a+b)+6\cdot 2b+8\cdot 3b$$ $$=4a+4(a+b)+12b+24b$$ $$=8a+40b$$ So, $$8a+40b=\frac{46}{9}$$ $$4a+20b=\frac{23}{9} \quad ...(2)$$ 3. **Solve for $a$ and $b$** From (1): $$2a+3b=\frac12$$ Multiply by $2$: $$4a+6b=1 \quad ...(3)$$ Subtract (3) from (2): $$ (4a+20b)-(4a+6b)=\frac{23}{9}-1$$ $$14b=\frac{14}{9}$$ $$b=\frac19$$ Now from (1): $$2a+3\left(\frac19\right)=\frac12$$ $$2a+\frac13=\frac12$$ $$2a=\frac16$$ $$a=\frac{1}{12}$$ 4. **Compute $E(X^2)$** $$E(X^2)=0^2\cdot a+2^2\cdot 2a+4^2(a+b)+6^2\cdot 2b+8^2\cdot 3b$$ $$=4\cdot 2a+16(a+b)+36\cdot 2b+64\cdot 3b$$ $$=8a+16a+16b+72b+192b$$ $$=24a+280b$$ Substitute $a=\frac{1}{12},\; b=\frac19$: $$E(X^2)=24\left(\frac{1}{12}\right)+280\left(\frac19\right)$$ $$=2+\frac{280}{9}$$ $$=\frac{18+280}{9}=\frac{298}{9}$$ 5. **Compute the variance** $$\operatorname{Var}(X)=E(X^2)-[E(X)]^2$$ $$=\frac{298}{9}-\left(\frac{46}{9}\right)^2$$ $$=\frac{298}{9}-\frac{2116}{81}$$ $$=\frac{2682-2116}{81}$$ $$=\frac{566}{81}$$ 6. **Match with options** $$\boxed{\frac{566}{81}}$$ So the correct option is **B**.