Question

A recent kellogg survey shows that approximately 34% of americans eat breakfast. suppose alex takes a random sample of four americans. what is the probability that at least one american eats breakfast or at least three americans do not eat breakfast?

Answer 1

This kind of experiments are ruled by Bernoulli's formula. If you have probability p of "success", and you want k successes in n trials, the probability is

$\binom{n}{k}p^k(1-p)^{n-k}$

It's easier to compute the first probability by difference: instead of computing the probability of the event "at least one of the surveyed eats breakfast", let's compute the probability of its contrary: none of them eats breakfast. So, we want 0 successes in 4 trials, with probability of success 0.34. The formula yields

$\binom{4}{0} 0.34^0 0.64^4 = 0.64^4 \approx 0.17 = 17\%$

Since the contrary has probability 17%, our event "at least one of the surveyed eats breakfast" has probability 83%.

As for the second question, the event "at least three of the surveyed eats breakfast" is the union of the events "exactly three of the surveyed eats breakfast" and "exactly four of the surveyed eats breakfast". So, we just need to sum their probabilities:

$\binom{4}{3} 0.34^3 0.64^1 +\binom{4}{4} 0.34^4 0.64^0 = 4\cdot 0.34^3 0.64 + 0.34^4 \approx 0.11 = 11\%$