Given that x is a hypergeometric random variable with Nequals10, nequals4, and requals6, complete parts a through d. a. Compute P(xequals1). P(xequals1)equals . 114 (Round to three decimal places as needed.) by. Compute P(x equals 0). P(xequals 0)equals . 005 (Rounded to three decimal places as needed.) c. Compute P(xequals3). P(xequals3)equals . 381 (Round to three decimal places as needed.) d. Compute P(xgreater than or equals5). P(xgreater than or equals5)equals nothing (Round to three decimal places as needed.)

a. 0.114

b. 0.005

c. 0.381

d. 0.00

Step-by-step explanation:

If x follows a hypergeometric distribution, the probability that x is equal to k is calculated as:

Where k≤r and n-k≤N-r, Additionally:

So, replacing N by 10, n by 4 and r by 6, we get:

Then, the probability that x is equal to 1, P(x=1) is:

The probability that x is equal to 0, P(x=0) is:

The probability that x is equal to 3, P(x=3) is:

Finally, in this case, x can take values from 0 to 4, so the probability that x is greater or equals to 5 is zero.

ian gone lie this got me confused

step-by-step explanation: