JEE Main 2026 — Mathematics Probability & Statistics

Let the outcomes of the $8$8 tosses be denoted by $X_1, X_2, X_3, X_4, X_5, X_6, X_7, X_8$X1​,X2​,X3​,X4​,X5​,X6​,X7​,X8​.

Let $A$A be the number of heads in the first $3$3 tosses ($X_1, X_2, X_3$X1​,X2​,X3​).

Let $B$B be the number of heads in the next $3$3 tosses ($X_4, X_5, X_6$X4​,X5​,X6​).

Let $C$C be the number of heads in the last $2$2 tosses ($X_7, X_8$X7​,X8​).

According to the given conditions:

Number of heads in the first six tosses: $A + B = 4$A+B=4

Number of heads in the last five tosses: $B + C = 3$B+C=3

Since $C$C is the number of heads in $2$2 tosses, $C \le 2$C≤2. Thus, $B = 3 - C \ge 1$B=3−C≥1. Also, $B \le 3$B≤3.

The possible values for $B$B are $1, 2, 3$1,2,3. We analyze each case:

Case 1: $B = 1$B=1

Then $C = 2$C=2 and $A = 3$A=3.

Number of ways = $^{3}C_{3} \times ^{3}C_{1} \times ^{2}C_{2} = 1 \times 3 \times 1 = 3$3C3​×3C1​×2C2​=1×3×1=3

Case 2: $B = 2$B=2

Then $C = 1$C=1 and $A = 2$A=2.

Number of ways = $^{3}C_{2} \times ^{3}C_{2} \times ^{2}C_{1} = 3 \times 3 \times 2 = 18$3C2​×3C2​×2C1​=3×3×2=18

Case 3: $B = 3$B=3

Then $C = 0$C=0 and $A = 1$A=1.

Number of ways = $^{3}C_{1} \times ^{3}C_{3} \times ^{2}C_{0} = 3 \times 1 \times 1 = 3$3C1​×3C3​×2C0​=3×1×1=3

Total number of favorable outcomes = $3 + 18 + 3 = 24$3+18+3=24

Total number of possible outcomes for $8$8 coin tosses = $2^8 = 256$28=256

The probability $p$p is given by:

$$p = \dfrac{24}{256} = \dfrac{3}{32}$$p=25624​=323​

We need to find the value of $96p$96p:

$$96p = 96 \times \dfrac{3}{32} = 3 \times 3 = 9$$96p=96×323​=3×3=9

Answer: $9$9