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