Maths with statistics HELP PLZ!!!!?

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 i am really stuck

Maths with statistics HELP PLZ!!!!?
a) P(B=1) = 2/6 = 1/3





P(B=2) = [1 - P(B=1)] x 2/5 = 1/3 x 2/5 = 4/15





b) P(B=3) = [1 - P(B=1) - P(B=2)] x 2/4


= [1 - 1/3 - 4/15] x 2/4 = 15/15- 5/15 - 4/15] x 2/4


= 6/15 x 2/4 = 12/60 = 1/5





P(B=4) = [1 - P(B=1) - P(B=2) - P(B=3)] x 2/3


= [1 - 1/3 - 4/15 - 1/5] x 2/3 = [15/15 - 5/15 - 4/15 -3/15] x 2/3


= 3/15 x 2/3 = 2/15





P(B=5) = 1/15





P(b%26gt;5) = 0





c) E(B) = 1x5/15 + 2x4/15 + 3x3/15 + 4x2/15 + 5x1/15


= 35/15 = 7/3 = 2.333333333333333








"if one of the first two balls picked out is a blue" is ambiguous





d) For exactly one blue:





P = [2/6 x 4/5] + [4/6 x 2/5] = 16/30 = 8/15


E = .5 x 8/15 = 4/15 = 2.66666666666666





e) E = 1 x 8/15 x 3 = 24/15 = 1.6





d) For at least one blue:





P = P(B=1) + P(B = 2) = 5/15 + 9/15 = 9/15





e) E = 1 x 9/15 x 3 = 27/15 = 1.8

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