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.
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%
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.
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%