SOLUTION:
Case: Binomial Probability
Method:
The formula for the probability is:
[tex]P=^nC_rp^rq^{n-r}[/tex]
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:
[tex]\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}[/tex]
Final answer:
The probability of obtaining at least 5 twos is: 0.002
