JEE Main 2026 — Probability Question with Solution

Let the outcomes of the $8$ tosses be denoted by $X_1,X_2,X_3,X_4,X_5,X_6,X_7,X_8$.

Let $A$ be the number of heads in the first $3$ tosses ($X_1,X_2,X_3$).
Let $B$ be the number of heads in the next $3$ tosses ($X_4,X_5,X_6$).
Let $C$ be the number of heads in the last $2$ tosses ($X_7,X_8$).

According to the given conditions:
Number of heads in the first six tosses: $A+B=4$
Number of heads in the last five tosses: $B+C=3$

Since $C$ is the number of heads in $2$ tosses, $C\le 2$. Thus, $B=3-C\ge 1$. Also, $B\le 3$.
The possible values for $B$ are $1,2,3$. We analyze each case:

Case 1: $B=1$
Then $C=2$ and $A=3$.
Number of ways = ${}^3{C}_3{\times }^3{C}_1{\times }^2{C}_2=1\times 3\times 1=3$

Case 2: $B=2$
Then $C=1$ and $A=2$.
Number of ways = ${}^3{C}_2{\times }^3{C}_2{\times }^2{C}_1=3\times 3\times 2=18$

Case 3: $B=3$
Then $C=0$ and $A=1$.
Number of ways = ${}^3{C}_1{\times }^3{C}_3{\times }^2{C}_0=3\times 1\times 1=3$

Total number of favorable outcomes = $3+18+3=24$
Total number of possible outcomes for $8$ coin tosses = $2^8=256$

The probability $p$ is given by:
$p=\frac{24}{256}=\frac{3}{32}$

We need to find the value of $96p$:
$96p=96\times \frac{3}{32}=3\times 3=9$

Answer: $9$