SOLUTION: the probability of getting "A" on your math final exam is 1/6 . the probability of an "A" on your English final exam 2/3. Find a) the probability of getting an "A" in math and a

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Question 1087910: the probability of getting "A" on your math final exam is 1/6 . the probability of an "A" on your English final exam 2/3. Find
a) the probability of getting an "A" in math and an "A" in english.
b) the probability of getting an "A" in math or an "A" in English.
Answer by jim_thompson5910(35256) (Show Source):
You can put this solution on YOUR website!
Let,
M = event that you get an "A" in math
E = event that you get an "A" in English

These two events are independent from one another

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Part a)

P(M) = probability you get an "A" in math (ie probability that event M occurs)
P(M) = 1/6

P(E) = probability you get an "A" in English (ie probability that event E occurs)
P(E) = 2/3

P(M and E) = probability you get an "A" in math, and an "A" in english (probability both events occur)
P(M and E) = P(M)*P(E) ... only works because M and E are independent events
P(M and E) = (1/6)*(2/3)
P(M and E) = (1*2)/(6*3)
P(M and E) = (1*2)/(18)
P(M and E) = (1*2)/(9*2)
P(M and E) = 1/9

The probability of getting an "A" in math and an "A" in English is 1/9 which is in fraction form (note: 1/9 = 0.111 = 11.1% approximately)

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Part b)

P(M or E) = P(M) + P(E) - P(M and E)
P(M or E) = (1/6) + (2/3) - (1/9)
P(M or E) = (1/6)*(3/3) + (2/3)*(6/6) - (1/9)*(2/2)
P(M or E) = (3/18) + (12/18) - (2/18)
P(M or E) = (3+12-2)/18
P(M or E) = 13/18

The probability of getting an "A" in math, or an "A" in English is the fraction 13/18 (note: 13/18 = 0.722 = 72.2% approximately)

The keyword "Or" here means that we can pick one event or the other, but not both at the same time.