Question

If a fair die is rolled 7 times, what is the probability, rounded to the nearest thousandth, of getting at least 5 twos?

Answer 1

SOLUTION:

Case: Binomial Probability

Method:

The formula for the probability is:

$P=^nC_rp^rq^{n-r}$

number of rolls, n= 7

required outcomes, r>= 5

prob of obtaining a 5, p= 1/6

prob of not obtaining a 5, q= 5/6

Hence:

$\begin{gathered} 7C5.(\frac{1}{6})^5(\frac{5}{6})^2+7C6.(\frac{1}{6})^6(\frac{5}{6})^1+7C7.(\frac{1}{6})^7(\frac{5}{6})^0 \\ \frac{21\times25}{6^7}+\frac{7\times5}{6^7}+\frac{1\times1}{6^7} \\ \frac{525+35+1}{6^7} \\ \frac{561}{6^7} \\ =0.002 \end{gathered}$

Final answer:

The probability of obtaining at least 5 twos is: 0.002