A fair die is tossed 4 times. Find the probability of a. at least 2 5s b. at most 3 even numbers c. no more than 1 odd number d. at least 1 prime number e. at most 1 number greater than 4

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A fair die is tossed 4 times. Find the probability of a. at least 2 5s b. at most 3 even numbers c. no more than 1 odd number d. at least 1 prime number e. at most 1 number greater than 4

Mathematics
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to find a), you need to know that P(at least 2 5's) = P(exactly 2 5s) + P(exactly 3 5s) +P(exactly 4 5s)+P(all 5s)
so you can find each of those probabilities and add them. that would be the straight forward way.
the easier way is to consider the complement of P(at least 5) which is P(at most 1)

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that way all you have to do is calculate P(at most 1 5s) = P(no 5) + P(exactly 1 5)
Then you can use the fact that P(event A) = 1 - P( complement of event A)
does that help you at all ?
From the looks of it, all problems I see here becomes much easier if you use P(A) = 1-P(notA) let me know if you still need help
how about the other parts?

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