fauji wrote:
The probability of getting a '6' on a role of die is 1/6. If this die is rolled
4 times, what is the probability of getting a '6' at least 2 times?
(A) \((1/6)^4\)
(B) \(2(1 /6)^3(5/6) + (1/6)^4\)
(C) \(6(1/6)^2(5/6)^2 + 4(1/6)^3(5/6) + (1/6)^4\)
(D) \(12(1/6)^2(5/6)^2 + 4(1/6)^3(5/6) + (1/6)^4\)
(E) \(12(1/6)^2(5/6)^2 + 6(1/6)^3(5/6) + (1/6)^4\)
Ok let us see the solution in a bit detail
The probability of getting a particular number is \(\frac{1}{6}\) and
NOT getting is \(\frac{5}{6}\).
1. 2 timesSo 2 times the probability will be \(\frac{1}{6}\) and other two times \(\frac{5}{6}\) as any of the 5 can be chosen
Now this 2 times can be chosen in 4C2 ways, so \(4C2*(\frac{1}{6})^2*(\frac{5}{6})^2=6*(\frac{1}{6})^2*(\frac{5}{6})^2\)
2. 3 times So 3 times the probability will be \(\frac{1}{6}\) and one times \(\frac{5}{6}\) as any of the 5 can be chosen
Now this 3 times can be chosen in 4C3 ways, so \(4C3*(\frac{1}{6})^3*(\frac{5}{6})^1=4*(\frac{1}{6})^3*(\frac{5}{6})\)
3. 4 times So all 4 times the probability will be \(\frac{1}{6}\).
Now this 4 times can be chosen in 4C4 ways, so \(4C4*(\frac{1}{6})^4=(\frac{1}{6})^4\\
\)
Add all 3 and you will get the answerC
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