A person has a 20 percent chance of winning on a scratch-off lottery ticket. What is the probability she first wins of the fourth ticket?

(Four over One) (0.20)3 (0.80)

(Four over Three) (0.80)3 (0.20)

(0.20)3 (0.80)

(0.80)3 (0.20)

0.20

Option B. (4 over 3)*(0.80)^3*0.20

Step-by-step explanation:

Given:

Probability of success p=0.20

Probability of failure q=0.80

To find:

Probability that she wins 1st of 4th ticket.

Solution:

By using binomial distribution formula,

P={nCr *p^k*q^(n-k)}

Total no of combination=nCk

Here n=4 and k=1

So,

total combination =4C1

=4!/(1!*3!)

=4 over 3 =4!/3!

So,

getting the probability that 1st win on fourth ticket with 0.20 % winning chance is

=p^k*q^(n-k)

=0.2^1*0.8(4-1)

=0.2*0.8^3

Hence The probability that she wins 1st that of the fourth ticket is

=(4 over 3)*(0.80)^3*0.20

the answer is stop cheating on things stoopid

step-by-step explanation:

im eatin up the 44