A coin is tossed 8 times.

Question

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 \_\_\_\_.

Answer: 9

Accepted Answer

Let the outcomes be $x_1,\dots,x_8$. The blocks $(x_1,\dots,x_6)$ and $(x_4,\dots,x_8)$ overlap in $(x_4,x_5,x_6)$. Let $k$ be the number of heads among $x_4,x_5,x_6$.

Then $(x_1,x_2,x_3)$ must contain $4-k$ heads, and $(x_7,x_8)$ must contain $3-k$ heads.

- $k=1$: ${}^3C_1$ (for the overlap) $\times\ {}^3C_3$ (3 heads in first block) $\times\ {}^2C_2$ (2 heads in last block) $= 3\times1\times1=3$
- $k=2$: ${}^3C_2 \times {}^3C_2 \times {}^2C_1 = 3\times3\times2=18$
- $k=3$: ${}^3C_3 \times {}^3C_1 \times {}^2C_0 = 1\times3\times1=3$

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

$$p = 24\left(\frac12\right)^8 = \frac{24}{256}=\frac{3}{32}$$

$$96p = 96\times\frac{3}{32}=9$$

Solutions · 1