Qa - binomial distributions - q2

1.np = 10/3.2*nc2*p^2*q^(n-2)= nc3*p^3*q^n-3 implies 2*(1-p) = ((n-2)/3)*p implies p =2/3 and n =5 now P(X<=2) = P(
X=0)+P(X=1)+P(X=2) = (1/3)^5 + 5*(2/3)*(1/3)^4 + 10*(2/3)^2*(1/3)^3 = (1+10+40)/243 = 51/243 = 17/81 answer is b

2. assume that  being a tea drinker is like taking a flip of coin.(i.e. either the person drinks tea or not) with the probability of heads being 1/3 and tails being 2/3 now P(X>=5) =P(X=5)+P(X=6)+P(X=7)+P(X=8) = (56*8+28*4+8*2+1)/6561 = 577/6561 =0.087943  so for 1000 trials the number of trials which report greater than or equal to 5 is 1000*P(X>=5) = 87.94 = 88 so answer is c

1.np = 10/3.2*nc2*p^2*q^(n-2)= nc3*p^3*q^n-3 implies 2*(1-p) = ((n-2)/3)*p implies p =2/3 and n =5 now P(X<=2) = P(
X=0)+P(X=1)+P(X=2) = (1/3)^5 + 5*(2/3)*(1/3)^4 + 10*(2/3)^2*(1/3)^3 = (1+10+40)/243 = 51/243 = 17/81 answer is b

2. assume that  being a tea drinker is like taking a flip of coin.(i.e. either the person drinks tea or not) with the probability of heads being 1/3 and tails being 2/3 now P(X>=5) =P(X=5)+P(X=6)+P(X=7)+P(X=8) = (56*8+28*4+8*2+1)/6561 = 577/6561 =0.087943  so for 1000 trials the number of trials which report greater than or equal to 5 is 1000*P(X>=5) = 87.94 = 88 so answer is c

ans is 51/243.

p( x less than equal to 2) = p(0) +p(1)+ p(2)

                                              =1/243+10/243+40/243

                                              =51/243

e(x) = 1/3(5+6+7+8)

       = 8.8

        =8.8*10

         = 88.

Thanks for the answers. Found out where i went wrong... Thanks.