suppose that purine is binomially distributed with probablit

suppose that purine is binomially distributed with probablity 0.7 for microRNA. given microRNA size 20

1) what is probablity of having 3 purines

2) what is proabability of having less than or equal to 3 purines?

3) what is probablity of having more than 4 purines

4) what is probablity of having number of purines between 10 and 15?

suppose that purine is binomially distributed with probablity 0.7 for microRNA. given microRNA size 20

The formula for the binomial probability is:

P(X = x) = nCx*p^x*(1-p)^(n-x)

Since n is 20, and p is .7:

P(X = x) = 20Cx*.7^x(.3)^(20-x)

1) what is probablity of having 3 purines

P(X = 3) = 20C3*.7^3(.3)^(20-3) = 0.0000005

Alternatively on a calculator with a binompdf function: binompdf(20, .7, 3) = 0.0000005

2) what is proabability of having less than or equal to 3 purines?

For less than or equal to 3, x can equal 0, 1, 2, or 3

P(X = 0) = 20C0*.7^0(.3)^(20-0) = 0.0000000000349

P(X = 1) = 20C1*.7^1(.3)^(20-1) = 0.0000000016272

P(X = 2) = 20C3*.7^2(.3)^(20-2) = 0.0000000360688

P(X = 3) = 20C3*.7^3(.3)^(20-3) = 0.000000505

Add these probabilities up: The sum is 0.0000005

Answer: 0.0000005

Alternatively, on a calculator with a binomcdf function: binomcdf(20, .7, 3) = 0.0000005

3) what is probablity of having more than 4 purines

The probability of more than 4 is the same as 1 minus the probability of 4 or less.

Four or less: On a calculator: binomcdf(20, .7, 4) = 0.0000056

So the probability of 4 or less is 0.0000056 and the probability of more than 4 is 1