If six people decide to come to a basketball game, but three of them are only 2/5 sure that they will stay for the entire time (the other three are sure they'll stay the whole time), what is the probability that at the end, at least 5 people stayed the entire time?

So far I have
(3/5)^3*1 is the possibilty everyone stays and then (3 choose 2)*(3/5)^2*(2/5) for only five ppl staying. Is this right?

Question

If six people decide to come to a basketball game, but three of them are only 2/5 sure that they will stay for the entire time (the other three are sure they'll stay the whole time), what is the probability that at the end, at least 5 people stayed the entire time?

So far I have
(3/5)^3*1 is the possibilty everyone stays and then (3 choose 2)*(3/5)^2*(2/5) for only five ppl staying. Is this right?

axie · Answer

@cripquezz

SmokeyBrown · Answer

I think the probability of everyone staying would be (2/5)^3, since the probability of each one of these 3 uncertain people staying is (2/5)
And then you'd consider the probability of only 2 of these 3 staying. My approach to this would be to consider 3 scenarios: person 1 and 2 stay; person 1 and 3 stay; person 2 and 3 stay. Each of these scenarios has a probability of (2/5)^2, so the total probability of "exactly 2 uncertain people stay" would be 3*(2/5)^2. I'm not sure if this would be the same result as your approach of (3 choose 2)*(3/5)^2*(2/5), and I'll admit my grasp of probability isn't the strongest, so it's quite possible I'm mistaken in my approach anyway...

In any case, if you can find the probability of all 3 uncertain people staying as well as the separate probability of any 2 of the 3 uncertain people staying, then I think you would ADD these two probabilities in order to find the probability of "at least 5 people staying"

crispyrat · Answer

I fixed it I had mixed it up it would be 2/5 where 3/5 is and vice versa

SmokeyBrown · Answer

Nice. Glad to hear you figured it out!