Can you please answer all parts (A,B & C). Thank you In a month with 24 working
ID: 3300974 • Letter: C
Question
Can you please answer all parts (A,B & C). Thank you
In a month with 24 working days, you decide to come to work on the first 12 working days by bike and the next 12 working days by walking. If you bike to work, the probability of you running into your friend Brad is 1/4, independent across days. If you walk to work, the probability of you running into your friend Brad is 2/3 again independent across days. For the problems below, leave expressions of the form "1/2^10" as is, i.e. you don't need to plug this into a calculator (you will get a very small number!). (a) What is the chance you run into Brad exactly once while you are biking? (b) What is the chance you run into Brad exactly twice when you are walking to work? c) What is the chance that the total number of times you run into your friend Brad in the month while going to work in the month is less than or equal to (lessthanorequalto) 2 times?Explanation / Answer
Let p1 is the probability of running into your friend Brad while bike to work, p1 = 1/4
and p2 is the probability of running into your friend Brad while walking to work, p2 = 2/3
(a)
Out of 12 days while biking to work exactly once you meet Brad
Probability of this = 12C1 * (1/4) * (1 - 1/4)^11
Probability = 12 * (1/4) * (3/4)^11 = 3*(3/4)^11
(b)
Out of 12 days while walking to work exactly once you meet Brad
Probability of this = 12C1 * (2/3) * (1 - 2/3)^11
Probability = 12 * (2/3) * (1/3)^11 = 8*(1/3)^11
(c)
Probability that not meeting Brad in a month = (1 - 1/4)^12 * (1 - 2/3)^12 = (3/4)^12 * (1/3)^12
Probability that meeting Brad in a month exactly once either while biking or walking (probability of meeting while biking and not meeting while walking OR probability of meeting while walking and not meeting while biking) = 3*(3/4)^11 * (1/3)^12 + 8*(1/3)^11 * (3/4)^12
Probability of meeting Brad exactly one day = 3*(3/4)^11 * (1/3)^12 + 8*(1/3)^11 * (3/4)^12
Meeting Brad on two days while biking = 12C2 * (1/4)^2 * (1-1/4)^10 = 66*(1/4)^2*(3/4)^10
Meeting Brad on two days while walking = 12C2 * (2/3)^2 * (1-2/3)^10 = 66*(2/3)^2*(1/3)^10
Probability of meeting Brad exactly 2 days = 3*(3/4)^11*8*(1/3)^11 + 66*(1/4)^2*(3/4)^10 + 66*(2/3)^2*(1/3)^10
Hence required probability = probability of not meeting Brad + probability of meeting Brad exactly 1 day + Probability of meeting Brad exactly two days