137 cards are drawn at random from a pack of 52 cards what is probability that 3 red 4 black P(A)=\frac{26}{52}\frac{25}{51}\frac{24}{50}\frac{26}{49}\frac{25}{48}\frac{24}{47}\frac{23}{46}=\frac{325}{39151}
i dunt know why i am getting wrong answer for so easy question so i will post my answetr and the co0rrect answer also
7 cards are drawn at random from a pack of 52 cards what is probability that 3 red 4 black
answer 325 /39151
my answer 26C3 .26C4 / 52C7 numerically bhi alag aate hai [2]
the followin question i had a doubt
so please solve them
A and B throw alternatevely with a pair of dice . A wins if he throws 6 before B throws 7 and B wins if he throws 7 before A 6 find the chances of winning if A makes the first throw
also find the chance that B will win .
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UP 0 DOWN 0 0 7
7 Answers
13
62spider..
you have to multiply the answer you are geting by 7! also..
You have just chosen them..
I think you have not taken the order into consideration.
13A and B throw alternatevely with a pair of dice . A wins if he throws 6 before B throws 7 and B wins if he throws 7 before A 6 find the chances of winning if A makes the first throw
Probability of winning of A=5/36+31/36*5/6*5/36+31/36*5/6*31/36*5/6*5/36+.........∞
62A wins if he throws 6 before B throws 7 and B wins if he throws 7 before A 6 find the chances of winning if A makes the first throw
Another way by reccursion.. see if you understand
p- probability that A throws 6 = 5/36
q - probabilty that B throws 7 = 6/36
P(A)= p+(1-p)(1-q)P(A)
P(A)= 5/36+31x5/216P(A)
P(A)= 5/36+155/216P(A)
61/216 P(A)= 5/36
P(A)= 30/61
P(B)=31/61
IS this proof obvious? And have i made any calculation mistakes.. pls check