A recent kellogg survey shows that approximately 34% of americans eat breakfast. suppose alex takes a random sample of four amer

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6 days ago

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A recent kellogg survey shows that approximately 34% of americans eat breakfast. suppose alex takes a random sample of four amer

icans. what is the probability that at least one american eats breakfast or at least three americans do not eat breakfast?

Mathematics

1 answer:

Svet_ta [4.3K] · Answer 1 · 2026-02-27T18:19:09+03:00

These experiments depend on Bernoulli's principle. If the likelihood of "success" is p and you are looking for k successes out of n attempts, the probability can be expressed as:

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

To simplify the calculation for the first probability, instead of determining the chances of at least one subject having breakfast, we calculate the opposite: that none do. Thus, we seek 0 successes from 4 trials with a success probability of 0.34. Using the formula gives us:

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

With the contrary outcome at 17%, the probability of our event "at least one of the surveyed eats breakfast" is 83%.

Regarding the second question, the event "at least three of the surveyed eat breakfast" encompasses the events of "exactly three eating breakfast" and "exactly four eating breakfast." So, we just need to add their probabilities:

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