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Mathematics: Post your doubts here!

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Eight cards are selected with replacement from a standard pack of playing cards
with 12 pic cards, 20 odd cards
and 20even cards.
a) how many different sequences of cards are possible?
b) how many of the sequencees in part (a) will contain three picture cards, three odd cars and two even cards??
 
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FIRST FIND THE PROBABILITY THAT RAJ's score is 12 with both even and odd numbers on the spinner.
Probability of getting an even number on the spinner is 2/5 and then of getting 12 is P(2,6)*2 + P(3,4)*2
So it becomes 2/5 *( P(2,6)*2 + P(3,4)*2)=2/45
now for the odd one. Probability of getting an odd number on the spinner is 3/5 but this time u will have to add two numbers to get 12 and hence p(6,6)=3/5*1/6*1/6=1/60
so P(SCORE=12) is 1/60 +2/45 =11/180
To find the conditional probability P(E|12) =2/45 devided by 11/180 = 8/11 :)
But where do you get the six from? Its five-sided?
 
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GUYS. come on. some help would be highly appreciated. i"ve posted so many queries..plz answer..paper tomorrow!!
 
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http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_61.pdf
Q3i..how d owe get the probability table? as in i dont knw how to get the probabilites..i dont understand ms method..
x 1 2 3 4 because there are 4 people to be selected and x represents the number of girls selected.
P(X=1) = p(bbbg)*4!/3!=3/8 * 2/7 *1/6 * 5/5 * 4= 1/14
P(X=2) =p(bbgg) *4!/(2!*2!) = 3/8 * 2/7 * 5/6 * 4/5 *6 = 3/7
P(X=3) =p(bggg) *4!/3! = 3/8 * 5/7 *4/6*3/5 *4 = 3/7
P(X=4) =p(gggg) *4!/4! = 5/8 * 4/7 * 3/6 * 2/5 = 1/14
You have to make arrangements here to because any guy/girl can be selected first .Here you go :p just remember after a girl/guy is selected he/she cant be selected again so consider without replacement
 
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I keep getting 7.61 :S
and question 4(ii) i found the LQ and UQ and got that right but i dont get how they got the 5 and 20 :/ sorry
just use more than 3 s.f for the mean which u take while calculating the standard deviation :) it gives a more precise answer
 
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