Question 1087910
<font color="black" face="times" size="3">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) = <font color=red size=4>1/9</font>


The probability of getting an "A" in math and an "A" in English is <font color=red size=4>1/9</font> 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) = <font color=red size=4>13/18</font>


The probability of getting an "A" in math, or an "A" in English is the fraction <font color=red size=4>13/18</font> (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. </font>