JEE Main 2022 — Mathematics Probability & Statistics

Given,

$A={1,2,3,4}:P(A)=\frac{3}{4}\rightarrow$A=1,2,3,4:P(A)=43​→Correct answer

$B={5,6,7,8,9,10};P(B)=\frac{1}{4}$B=5,6,7,8,9,10;P(B)=41​ Correct answer

Now total ways in which $8$8 Correct answer is given will be,

$$\Rightarrow \text{Case (i) }(4,4):C44{(\frac{3}{4})}^{4}\cdot C46\cdot {(\frac{1}{4})}^{4}\cdot {(\frac{3}{4})}^{2}$$⇒Case (i) (4,4):C44(43​)4⋅C46⋅(41​)4⋅(43​)2

$$\Rightarrow \mathrm{Case}\text{(ii) }(3,5):C34{(\frac{3}{4})}^{3}\cdot {(\frac{1}{4})}^{1}\cdot C56{(\frac{1}{4})}^{5}\cdot (\frac{3}{4})$$⇒Case(ii) (3,5):C34(43​)3⋅(41​)1⋅C56(41​)5⋅(43​)

$$\Rightarrow \text{Case (iii)}(2,6):C24{(\frac{3}{4})}^{2}{(\frac{1}{4})}^{2}\cdot C66{(\frac{1}{4})}^{6}$$⇒Case (iii)(2,6):C24(43​)2(41​)2⋅C66(41​)6

Total probability $=\frac{1}{{4}^{10}}[{3}^{4}\times 15\times {3}^{2}+4\times {3}^{3}\times 6\times 3+6\times {3}^{2}]$=4101​[34×15×32+4×33×6×3+6×32]

$$=\frac{27}{{4}^{10}}[2.7\times 15+72+2]$$=41027​[2.7×15+72+2]

Now comparing with $\frac{27k}{{4}^{10}}$41027k​

We get,$K=479$K=479