Sixty-six percent of US adults have very little confidence in newspapers. Ten adults are selected at random. Find the probability that the number with very little confidence is (a) exactly five, (b) at least six, and (c) fewer than four.

Step-by-step solution

1. 1

   Check that the binomial model applies. There are $n=10$ independent selections, each either 'has very little confidence' or not, with a constant probability $p=0.66$. So $X\sim B(10,0.66)$ and

   $P(X=k)=\binom{10}{k}(0.66)^{k}(0.34)^{10-k}.$

2. 2

   Part (a): exactly five.

   $P(5)=\binom{10}{5}(0.66)^{5}(0.34)^{5}=252(0.12523)(0.00454)=0.143.$

3. 3

   Part (b): at least six. Sum the terms from $k=6$ to $k=10$:

   $P(X\ge 6)=0.2331+0.2646+0.1991+0.0859+0.0157=0.773.$

   This is the larger tail because the mean $np=6.6$ sits inside it.

4. 4

   Part (c): decide whether to use the complement. 'Less than four' means $k=0,1,2,3$ — four terms. The complement $P(X\ge 4)$ would need seven terms, so here the direct sum is the shorter route. The complement is only a shortcut when the event you want has more cases than the one you do not.

5. 5

   Compute the four terms.

   $P(0)=(0.34)^{10}=0.0000206,\qquad P(1)=10(0.66)(0.34)^{9}=0.0004003,$
   $P(2)=45(0.66)^{2}(0.34)^{8}=0.0034970,\qquad P(3)=120(0.66)^{3}(0.34)^{7}=0.0180843.$

6. 6

   Add and round.

   $P(X<4)=0.0000206+0.0004003+0.0034970+0.0180843=0.02204\approx 0.022.$

   This is small, as expected: with a mean of $6.6$ successes, seeing three or fewer is well out in the lower tail.