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hi sorry, can someone give me abit of help of this question..

a) the random variable W has a B(6, 0.4) distribution. calculate the probability that the value of W is an odd number.

i made a p.d.f but when i added them up it didnt = 1, sumthing like 0.95..

???

b) the random variable X has a B(5, p) distribution, where p ≠ 0. given that E(X) = 3Var(X), find P(X=0)

c) the random variable Y has a B(12,p) distribution where p ≠ 0. given that

P(Y=11) = P(Y=12) find the vaule of p

if any1 can give me any hints or something then thankyou!

a) the random variable W has a B(6, 0.4) distribution. calculate the probability that the value of W is an odd number.

i made a p.d.f but when i added them up it didnt = 1, sumthing like 0.95..

???

b) the random variable X has a B(5, p) distribution, where p ≠ 0. given that E(X) = 3Var(X), find P(X=0)

c) the random variable Y has a B(12,p) distribution where p ≠ 0. given that

P(Y=11) = P(Y=12) find the vaule of p

if any1 can give me any hints or something then thankyou!

.x.g.x.

c) the random variable Y has a B(12,p) distribution where p ≠ 0. given that

P(Y=11) = P(Y=12) find the vaule of p

P(Y=11) = P(Y=12) find the vaule of p

P(Y = 11) = 12C11.(p)^11.(1-p)

P(Y = 12) = 12C12.(p)^12

P(Y = 11) = 12(p)^11.(1-p)

P(Y = 12) = (p)^12

(p)^12 = 12(p)^11.(1-p)

(p)^12 = 12(p)^11 - 12(p)^12

0 = 12(p)^11 - 13(p)^12

0 = 12/13p^11 - p^12

0 = p^11(12/13 - p)

0 = 12/13 - p

p = 12/13

--------------

.x.g.x.

hi sorry, can someone give me abit of help of this question..

a) the random variable W has a B(6, 0.4) distribution. calculate the probability that the value of W is an odd number.

i made a p.d.f but when i added them up it didnt = 1, sumthing like 0.95..

???

a) the random variable W has a B(6, 0.4) distribution. calculate the probability that the value of W is an odd number.

i made a p.d.f but when i added them up it didnt = 1, sumthing like 0.95..

???

Hehe! What is a p.d.f?

I haven't done further stats ...

(a)

P(W = 1) + P(W = 3) + P(W = 5)

= 6C1 (0.4)(0.6)^5 + 6C3 (0.4)^3(0.6)^3 + 6C5 (0.4)^5(0.6)

= 0.500

(b)

E(X) = 5p

Var(X) = 5p(1 - p)

Since E(X) = 3Var(X),

5p = 15p(1 - p)

p = 3p(1 - p)

p = 3p - 3p^2

3p^2 - 2p = 0

p(3p - 2) = 0

p = 2/3 . . . . . since p $\neq$ 0

P(X = 0)

= (1/3)^5

= 0.004

(c)

P(Y = 11)

= 12C11 p^11 (1 - p)

= 12 p^11 (1 - p)

P(Y = 12) = p^12

p^12 = 12 p^11 (1 - p)

p^11(p - 12(1 - p)) = 0

p^11(13p - 12) = 0

p = 12/13 . . . . . since p $\neq$ 0

P(W = 1) + P(W = 3) + P(W = 5)

= 6C1 (0.4)(0.6)^5 + 6C3 (0.4)^3(0.6)^3 + 6C5 (0.4)^5(0.6)

= 0.500

(b)

E(X) = 5p

Var(X) = 5p(1 - p)

Since E(X) = 3Var(X),

5p = 15p(1 - p)

p = 3p(1 - p)

p = 3p - 3p^2

3p^2 - 2p = 0

p(3p - 2) = 0

p = 2/3 . . . . . since p $\neq$ 0

P(X = 0)

= (1/3)^5

= 0.004

(c)

P(Y = 11)

= 12C11 p^11 (1 - p)

= 12 p^11 (1 - p)

P(Y = 12) = p^12

p^12 = 12 p^11 (1 - p)

p^11(p - 12(1 - p)) = 0

p^11(13p - 12) = 0

p = 12/13 . . . . . since p $\neq$ 0

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