It has been shown that 80% of a certain group of people prefer tea over coffee. Find the probability that, among eleven of these people, at least 7 prefer tea over coffee.

The probability that, among eleven of these people, at least 7 prefer tea over coffee is ___. (Round to three decimal places as needed.)


Answer:
Given:

Probability of success $(p)=0.80$

Sample size $(n)=11$

Solution:
The probability that among eleven of these people at least 7 prefer tea over coffee:
$$\text{P}(x\ge 7)=\sum _7^{11}(\genfrac{}{}{0ex}{}{n}{x})\cdot p^x\cdot (1-p{)}^{n-x}$$ $$=\sum _7^{11}(\genfrac{}{}{0ex}{}{11}{x})\cdot {0.80}^x\cdot (1-0.80{)}^{11-x}$$ $$=0.9496$$

Final Answer:
The probability that among eleven of these people at least 7 prefer tea over coffee $=0.9496$