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is it wrong to say E(X)= 0*(2/6) + 1*(8/30) +2*(24/120)..... and get 4/3
i dont know why the answer shows ... E(X)=0*(4/3) + 1*(8/30) + 2*(24/130)..... and ans shows 4/3 i got the same answer as the book states, however my expected value for 0 is different to the book.
fyi that prob distribution table was given in the question x 0 1 2 3 4 P(X=x) 2/6 8/30 24/120 2/15 24/360
your method is correct! :) there could be a typing error in the question or in the answer in your book. luckily, the answer didn't change because the erroneous number is multiplied to 0 :P
yes true. but while i have you here, could you please explain a bit more .. so table shows number of cards that need to be chosen before a queen appears .. if 0 cards are chosen - does that mean Q Q Q Q Q' Q'
can u explain how they got 2/6 for 0 in the table please
sure, x : number of cards chosen until the first queen appears x= 0 means the very first card chosen is a Queen! so what is the probability to choose a Queen from 2 Queens and 4 Kings?? total = 2+4 = 6 probability = 2/6 got it ?
do you also want an explanation how we get 8/30 for x =1 ?
x : number of cards chosen until the first queen appears x= 1 means the second card chosen is a Queen! which means that the first card chosen MUST be the king. so what is the probability to choose a King first "AND" then a Queen from (2 Queens and 4 Kings)?? [remember we multiply probabilities for AND and add for OR] probability of choosing a King from (2 Queens and 4 Kings) = 4/6 now there are only 5 cards remaining, king is already chosen, (2 Queens and 3 Kings) now we choose a Queen from these, probability = 2/5 (2 Queens from total 5 cards) so total probability = (4/6)* (2/5) = 8/30
oh right, i have started to draw a tree. and starting to get all the other values too and that is also helping me. I really appreciate your time. thank you so much.
i sort of misread the question and thought any 6 cards out of 52 cards, that why i was writing funny stuff at the top lol
ohh! :P it happens :) welcome ^_^