Probability: Let us suppose that each time a particular basketball player shoots a three point attempt the probability of scoring is 0.4. How many three point attempts does this player need to make for the probability of at least 3 successes to exceed 0.75
P(E)= at least 3 successes in n throws (n three point attempt) > 0.75 P(E')= maximum 2 successes in n throws (n three point attempts) P(E)+P(E')=1
I'd recommend to read P(E') as max 2 successes in n throws to be equal as exactly 0 success in n throws, exactly 1 success in n throws, exactly 2 successes in n throws. That's a binomial distribution leading to the Bernoulli Probability Distribution for a regular Bernoulli Trial. So sub divide P(E') into 3 events and compute the Bernoulli trial for each of them.
Thanks very much mate! worked out in the end. Thankyou for the help!