A die is thrown 6 times. Getting a number greater than 4 is considered a success. Find the probability of at least two successes.

Solution

Here n = 6
Let p be the probability of getting 5 or 6
$therefore space space space straight p space equals space straight P left parenthesis 5 space or space 6 right parenthesis space equals space 1 over 6 plus 1 over 6 space equals 2 over 6 space equals 1 third comma space space straight q space equals space 1 minus straight p space equals space 1 minus 1 third equals space 2 over 3$
P(at least two successes) = 1 - P(0) + P(1)
$equals space 1 minus open square brackets straight C presuperscript 6 subscript 0 open parentheses 2 over 3 close parentheses to the power of 6 space plus space straight C presuperscript 6 subscript 1 space open parentheses 2 over 3 close parentheses to the power of 5 space open parentheses 1 third close parentheses close square brackets space equals space 1 minus space open square brackets 1 space cross times 64 over 729 plus 192 over 729 close square brackets space equals 1 minus fraction numerator 64 plus 192 over denominator 729 end fraction equals space 1 minus 256 over 729 space equals 473 over 729 space$

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Probability

A die is thrown 6 times. Getting a number greater than 4 is considered a success. Find the probability of at least two successes.

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