Question

The probability of Andy and Anna having lunch together is 40%. The probability of them having lunch together at least 6 days a week is and the probability of having lunch exactly 6 times in a week is

Answer 1

The probability of having lunch together is p = 40% = 0.4
The probability of not having lunch together is q = 1 - p = 0.6

Number of trials (days in a week) is n = 7
Let r = number of days in the week when Andy and Anna have lunch together.
$P(r\,of\,n)=_{n}C_{r}\,p^{r}q^{n-r}$

Use th graphing calculator to obtain
P(6 of 7) = ₇C₆ (0.4)⁶(0.6) = 0.017
P(7 of 7) = ₇C₇ (0.4)⁷(0.6)⁰ = 0.002


Therefore
P(at least 6 of 7) = P(1 of 7) + P(2 of 7) + ... + P(6 of 7)
 = 0.131 + 0.261 + 0.290 + 0.194 + 0.077 + 0.017 
 = 0.97 or 97%

P(at least 6 of 7) = 0.017 + 0.002 = 0.019 = 1.9%
P(exactly 6 of 7) = 0.017 or 1.7%

Answer:
The probability of having lunch at least 6 days per week is 0.019 or  1.9%.
The probability of having lunch exactly 6 times is 0.017 or 1.7%