JEE Main 2023 — Mathematics Probability & Statistics

Given that variance of $\alpha -\beta$α−β is $\frac{p}{q}$qp​

$$\alpha \in {1,2,3,4,5,6}$$α∈1,2,3,4,5,6

$$\beta \in {1,2,3,4,5,6}$$β∈1,2,3,4,5,6

$(\alpha -\beta )=0$(α−β)=0 ($6$6 case)

$(\alpha -\beta )=-1$(α−β)=−1 ($5$5 case)

$(\alpha -\beta )=-2$(α−β)=−2 ($4$4 case)

$(\alpha -\beta )=-3$(α−β)=−3 ($3$3 case)

$(\alpha -\beta )=-4$(α−β)=−4 ($2$2 case)

$(\alpha -\beta )=-5$(α−β)=−5 ($1$1 case)

$(\alpha -\beta )=1$(α−β)=1 ($5$5 case)

$(\alpha -\beta )=2$(α−β)=2 ($4$4 case)

$(\alpha -\beta )=3$(α−β)=3 ($3$3 case)

$(\alpha -\beta )=4$(α−β)=4 ($2$2 case)

$(\alpha -\beta )=5$(α−β)=5 ($1$1 case)

Mean $=0$=0

We know that variance $\displaystyle =\frac{\sum _{i=1}^{n}{f}_{i}{({x}_{i}-x)}^{2}}{\sum _{i=1}^{n}{f}_{i}}$=∑i=1n​fi​∑i=1n​fi​(xi​−x)2​

Variance$={\sigma }^{2}=\frac{{0}^{2}\times 6+2\times {1}^{2}\times 5+2\times {2}^{2}\times 4+2\times {3}^{2}\times 3+2\times {4}^{2}\times 2\times {5}^{2}\times 1}{36}$=σ2=3602×6+2×12×5+2×22×4+2×32×3+2×42×2×52×1​

$$=\frac{2}{36}\times (5+16+27+32+25)=\frac{105}{18}=\frac{35}{6}$$=362​×(5+16+27+32+25)=18105​=635​

$$\therefore p=35$$∴p=35

Sum of divisors of $p=1+5+7+35=48$p=1+5+7+35=48

Option $(3)$(3) is correct.