Asked by Kennedy
two cards are dealt in succession from a standard deck of cards. what is the probability that the second card is red give that the first card was a heart. so i do p(R/H), then i know you go R*H/H. but how do i get the probability please explain! thanks.
if 2 dice are rolled, what is the probability of rolling at least one three?
i nkow that i have to use binomial probablity but how do i get the answer?
how do i simplify (x-2)!/x! i know that (x-2)! is (x-2)(x-1) but what is x! factorial? thanks for your time!
if 2 dice are rolled, what is the probability of rolling at least one three?
i nkow that i have to use binomial probablity but how do i get the answer?
how do i simplify (x-2)!/x! i know that (x-2)! is (x-2)(x-1) but what is x! factorial? thanks for your time!
Answers
Answered by
Damon
first card was red
so now I have 51 cards
25 are red and 26 are black
so
25/51
( are you sure you typed this question the exact way it was in the book?)
-------------------------------------
probability of the first 3 = 1/6 and the second not 3 = 1/6*5/6 = 5/36
since probability that the first was not 3 =5/6
probability that the second was 3 and the first not 3 = 1/6*5/6 = 5/36
probability that both are 3 = 1/6*1/6
sum = 1/6 + 5/36 + 5/36 = 12/36 = 11/36
OK, using BINOMIAL
p = 1/6
1-p = 5/6
P(n,k) = C(n,k) p^k (1-p)^(n-k)
we want P(2,1) + P(2,1)
C(2,1) = 2
C(2,2) = 1 from formula or Pascal triangle
P(2,1) = 2*(1/6)^1 *(5/6)^1
P(2,2) = 1*(1/6)^2 *(5/6)^0
SUM = 2*5/36 + 1/36 =11/36
so now I have 51 cards
25 are red and 26 are black
so
25/51
( are you sure you typed this question the exact way it was in the book?)
-------------------------------------
probability of the first 3 = 1/6 and the second not 3 = 1/6*5/6 = 5/36
since probability that the first was not 3 =5/6
probability that the second was 3 and the first not 3 = 1/6*5/6 = 5/36
probability that both are 3 = 1/6*1/6
sum = 1/6 + 5/36 + 5/36 = 12/36 = 11/36
OK, using BINOMIAL
p = 1/6
1-p = 5/6
P(n,k) = C(n,k) p^k (1-p)^(n-k)
we want P(2,1) + P(2,1)
C(2,1) = 2
C(2,2) = 1 from formula or Pascal triangle
P(2,1) = 2*(1/6)^1 *(5/6)^1
P(2,2) = 1*(1/6)^2 *(5/6)^0
SUM = 2*5/36 + 1/36 =11/36
Answered by
Damon
I did it both ways so you can study what is really going on.
Answered by
Damon
probability of the first 3 = 1/6 and the second not 3 = 1/6*5/6 = 5/36
since probability that the second was not 3 =5/6
probability that the second was 3 and the first not 3 = 1/6*5/6 = 5/36
probability that both are 3 = 1/6*1/6
sum = 1/6 + 5/36 + 5/36 = 11/36
OK, using BINOMIAL
p = 1/6
1-p = 5/6
P(n,k) = C(n,k) p^k (1-p)^(n-k)
we want P(2,1) + P(2,1)
C(2,1) = 2
C(2,2) = 1 from formula or Pascal triangle
P(2,1) = 2*(1/6)^1 *(5/6)^1
P(2,2) = 1*(1/6)^2 *(5/6)^0
SUM = 2*5/36 + 1/36 =11/36
since probability that the second was not 3 =5/6
probability that the second was 3 and the first not 3 = 1/6*5/6 = 5/36
probability that both are 3 = 1/6*1/6
sum = 1/6 + 5/36 + 5/36 = 11/36
OK, using BINOMIAL
p = 1/6
1-p = 5/6
P(n,k) = C(n,k) p^k (1-p)^(n-k)
we want P(2,1) + P(2,1)
C(2,1) = 2
C(2,2) = 1 from formula or Pascal triangle
P(2,1) = 2*(1/6)^1 *(5/6)^1
P(2,2) = 1*(1/6)^2 *(5/6)^0
SUM = 2*5/36 + 1/36 =11/36
Answered by
Damon
(x-2)!/x! i know that (x-2)! is (x-2)(x-1)
(x-2)(x-3)((x-4) .....
------------------------
x (x-1)(x-2)(x-3) (x-4) .....
= 1/ [ x(x-1)]
(x-2)(x-3)((x-4) .....
------------------------
x (x-1)(x-2)(x-3) (x-4) .....
= 1/ [ x(x-1)]
Answered by
Kennedy
i still don't under the simplifying one cause reiny told me that (x+2)! was (x+2) (x+1) but howcome you are telling me that you go up in factorial. i though factorial goes down
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