A local newspaper is holding a pun contest, where the chance of winning any priz

Question

A local newspaper is holding a pun contest, where the chance of winning any prize in the contest is 6%. What is the probability that if you send in 10 independent entries, hoping at least one of them would win, that no pun in ten did?
My question is why I can’t use formula p(at least one win)=1-p(no win)^10? A local newspaper is holding a pun contest, where the chance of winning any prize in the contest is 6%. What is the probability that if you send in 10 independent entries, hoping at least one of them would win, that no pun in ten did?
My question is why I can’t use formula p(at least one win)=1-p(no win)^10?
My question is why I can’t use formula p(at least one win)=1-p(no win)^10?

Explanation / Answer

Binomial distribution

n = 10

p = probability of success = 0.06

q = probability of failure = 1 - p = 0.94

To find P(no success) = P(X =0) = qn = 0.9410 = 0.5386

EXPLNATION:

Your question: Why I can't use the formula:
P(at least one win) = 1 - P(no win)10

My Answer: You cannot use that formula.

REASON: If you read the question asked very carefully:
You are not asked the probability of at least one win at all!!

The sentence is

You send 10 independent entries, hoping at least one of them would win, that no pun in ten did.

Here: To be noted is:

You are only hoping that at least one of them would win.

But, the question asked is:

the last 5 words: "that no pun in ten did".i.e., you are asked to calculate the probability that no pun did win.i.e., all are failures.

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