Question

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

Answer 1

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)

Answer 2

so you can find each of those probabilities and add them.
that would be the straight forward way.

Answer 3

the easier way is to consider the complement of P(at least 5)
which is P(at most 1)

Answer 4

that way all you have to do is calculate
P(at most 1 5s) = P(no 5) + P(exactly 1 5)

Answer 5

Then you can use the fact that
P(event A) = 1 - P( complement of event A)

Answer 6

does that help you at all ?

Answer 7

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

Answer 8

how about the other parts?