Fifty percent of consumers prefer to buy electronics online. You randomly select 8 consumers. Find the probability that the number who prefer to buy online is (a) exactly five, (b) more than five, and (c) at most five.

Step-by-step solution

1. 1

   Identify the distribution. Eight independent consumers, each either preferring online or not, with a fixed probability $p = 0.5$: this is binomial with $n = 8$, $p = 0.5$. The probability of exactly $k$ successes is

   $P(k) = \binom{8}{k}(0.5)^k(0.5)^{8-k} = \binom{8}{k}\frac{1}{256}$

   Because $p = 0.5$, the powers collapse to the single constant $\tfrac{1}{2^8} = \tfrac{1}{256}$, so the whole problem reduces to counting.

2. 2

   (a) Exactly five. $\binom{8}{5} = \dfrac{8!}{5!\,3!} = 56$, so

   $P(5) = \frac{56}{256} = 0.21875 \approx 0.219$

3. 3

   (b) More than five. "More than five" means $6$, $7$ or $8$, so add three terms. $\binom{8}{6} = 28$, $\binom{8}{7} = 8$, $\binom{8}{8} = 1$:

   $P(X > 5) = \frac{28 + 8 + 1}{256} = \frac{37}{256} = 0.144531 \approx 0.145$

   Note that $X > 5$ excludes $5$ — including it by mistake would add $0.219$.

4. 4

   (c) At most five, via the complement. "At most five" means $X \le 5$, which is everything except $X > 5$:

   $P(X \le 5) = 1 - P(X > 5) = 1 - \frac{37}{256} = \frac{219}{256} = 0.855469 \approx 0.855$

   This is far quicker than summing the six terms $k = 0,\ldots,5$.

5. 5

   Check that the pieces are consistent. Summing all $2^8$ outcomes: $\binom{8}{0}+\cdots+\binom{8}{8} = 256$, so the probabilities total exactly $1$ ✓. Also $P(X \le 5) + P(X > 5) = \tfrac{219}{256} + \tfrac{37}{256} = 1$ ✓, and $\tfrac{219}{256}$ was confirmed by directly summing $1+8+28+56+70+56 = 219$.

6. 6

   Sanity-check the shape. With $p = 0.5$ the distribution is symmetric about the mean $np = 4$, so $P(5)$ should equal $P(3)$: indeed $\binom{8}{3} = 56$ as well ✓. And since more than half the mass sits at or below $4$, a value of $0.855$ for $P(X \le 5)$ is exactly the right size.