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1 November, 16:38

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

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  1. 1 November, 17:54
    0
    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
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