Fifty percent of consumers prefer to buy electronics online. Eight consumers are chosen at random. 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

   Set up the binomial model and exploit p = 0.5. Here $X\sim B(8,0.5)$ and

   $P(X=k)=\binom{8}{k}(0.5)^{k}(0.5)^{8-k}=\frac{\binom{8}{k}}{2^{8}}=\frac{\binom{8}{k}}{256}.$

   Because $p=0.5$, the powers of $p$ and $1-p$ merge into a single $2^{-8}$, so every probability is just a binomial coefficient over $256$.

2. 2

   Write down the relevant coefficients. Row $8$ of Pascal's triangle is

   $1,\ 8,\ 28,\ 56,\ 70,\ 56,\ 28,\ 8,\ 1,$

   summing to $256$ as it must.

3. 3

   (a) Exactly five.

   $P(5)=\frac{\binom{8}{5}}{256}=\frac{56}{256}=0.21875\approx 0.219.$

4. 4

   (b) More than five. 'More than five' excludes five itself, so add $k=6,7,8$:

   $P(X>5)=\frac{28+8+1}{256}=\frac{37}{256}=0.14453\approx 0.145.$

   The boundary word matters: including $k=5$ here would give $0.363$ instead.

5. 5

   (c) At most five, using the complement. 'At most five' is everything except 'more than five', so no new sum is needed:

   $P(X\le 5)=1-P(X>5)=1-\frac{37}{256}=\frac{219}{256}=0.85547\approx 0.855.$

6. 6

   Check the arithmetic and the symmetry. The two complementary answers add to $0.145+0.855=1.000$ ✓. Also, $p=0.5$ makes the distribution symmetric about the mean $np=4$, so $P(X=5)=P(X=3)$ and $P(X>5)=P(X<3)$ — a quick way to spot a slipped coefficient.