Maths with Statistics (please help)?

A bag contains 4 red and 2 blue balls, all of the same size. A ball is slected at random and removed from the bag. This s repeated until a blue ball is removed from the bag.





The random variable B is the number of balls that have been removed from the bag





a) show that P(B=2) = 4/15


b) Find the probability distribution of B


c) Find E(B)





The bag and the same ball 6 balls are used in a game at a funfair. One ball is removed from the bag at a time and a contestant wins 50 pence (or cents for americans :D) if one of the first two balls picked out is a blue





d) What are the expected winnings of playing this game once





For 1 pound (or again 1 dollar), the contestant gets to play the game three times





e) What is the expected profit or loss from the three games





I know its long but please help me

Maths with Statistics (please help)?
a)


P(B=2) = prob a red is picked first, then a blue is picked


= (4/6) * (2/5) = 4/15





b)


P(B = b) = 4! / (5-b)! * (7-b)!/6! * 2/(7-b)





b P(B=b)


1 5/15


2 4/15


3 3/15


4 2/15


5 1/15





B can only be between 1 and 5 because there after 4 reds a picked, there's 100% probability that a blue is next.





c)


E(B) = 1*5/15 + 2*4/15 + 3*3/15 + 4*2/15 + 5*1/15 = 7/3 = 2.3333





d)


You win $0.50 if B%26lt;=2, and win $0 if B%26gt;2


P(B%26lt;=2) = 5/15 + 4/15 = 9/15 = .6


P(B%26gt;2) = 3/15 + 2/15 + 1/15 = 6/15 = .4





Expected winnings = .6*$0.50 + .4*$0 = $0.30





e)


The three games are independent of each other, so the expected winnings is $0.30 + $0.30 + $0.30 = $0.90





but you payed $1 to play, so the expected profit(loss) = $0.90 - $1.00 = -$0.10





Ten cent loss.