# Probability question

The Night Raiders, an expansion team in the National Indoor Football League, is having a challenging first season with a current win loss record of 0 and 4. However, the team recently signed four new defensive players and one of the team sponsors (who also happens to hold a CFA charter) calculates the probability of the team winning a game at 0.40. Assuming that whether the team wins a game is independent of whether it wins any other game, the probability that the team will win 6 out of the next 10 games is closest to: A) 0.112. B) 0.417. C) 0.350. D) 0.033.

Answer is: A) 0.112 p(x=6) = 10!/6!*4!*(0.4^6)*(0.6^4) = 0.112

I think rado is right A

Is this real exam level? Can’t imagine.

You mean you it’s so easy question like for the real exam level?

This is the formula of Binomial theoram.

Yes, there is no difficulty. Consider the following question (which would be easy, too, but a little bit more challenging for some candidates) Paul has probability to hit a target with his gun of 5 %. His shots are independant and have all the same prob to hit the target. How often does he have to shoot to hit at least on time with a total probability of at least 30 % A) 6 times B) 7 times C) 8 times D) 9 times Anybody in the game?

either my english is very bad, or the above question doesn’t make sense to me. can you re-phrase the last line? i hope its ‘one time’ and not ‘on time’.

the correct answer is 7 (B). probability of hitting a target after n shots is 1-.95^n = .3 n = log(.7)/log(.95) = 6.9536 1-(.95)^7 = .301663

It’s “one” time and maratikus has the solution I guess he likes math. he should be called mathikus not maratikus.