A coin is tossed 8 times. If the probability that exactly 4 heads appear in the first six tosses and exactly 3 heads appear in the last five tosses is $$p$$, then $$96p$$ is equal to _____.

Label the eight tosses as 1 2 3 4 5 6 7 8.

Two conditions are imposed:
  • Exactly 4 heads in the first six tosses (positions 1-6).
  • Exactly 3 heads in the last five tosses (positions 4-8).

Split the eight positions into three non-overlapping blocks:
  A: positions 1,2,3 (size 3)
  B: positions 4,5,6 (size 3)
  C: positions 7,8    (size 2)

Let $$a,b,c$$ denote the number of heads in blocks A, B, C respectively.

Translate the two conditions:
  • Heads in positions 1-6 ⇒ $$a+b=4$$  $$-(1)$$
  • Heads in positions 4-8 ⇒ $$b+c=3$$  $$-(2)$$

Because each block size limits its heads:
  0 ≤ $$a$$ ≤ 3,  0 ≤ $$b$$ ≤ 3,  0 ≤ $$c$$ ≤ 2.

Solving $$(1),(2)$$ simultaneously:

From $$(1):\;a=4-b$$ (so $$b=1,2,3$$).
From $$(2):\;c=3-b$$.

The admissible triples $$(a,b,c)$$ are therefore:

Case 1: $$b=1 \;\Rightarrow\; a=3,\;c=2$$ (possible)
Case 2: $$b=2 \;\Rightarrow\; a=2,\;c=1$$ (possible)
Case 3: $$b=3 \;\Rightarrow\; a=1,\;c=0$$ (possible)

Count sequences for each case using combinations.

Case 1: $$C(3,3)\,C(3,1)\,C(2,2)=1\cdot3\cdot1=3$$ sequences.

Case 2: $$C(3,2)\,C(3,2)\,C(2,1)=3\cdot3\cdot2=18$$ sequences.

Case 3: $$C(3,1)\,C(3,3)\,C(2,0)=3\cdot1\cdot1=3$$ sequences.

Total favourable sequences $$=3+18+3=24$$.

Total possible sequences of 8 tosses $$=2^8=256$$.

Hence $$p=\frac{24}{256}=\frac{3}{32}$$.

Required value: $$96p = 96 \times \frac{3}{32}=9$$.

Therefore, the answer is 9.