Answer:
Because there are 808080 total keys and 606060 open a classroom, we can say that the probability that a key opens a classroom, P(C)P(C)P, left parenthesis, C, right parenthesis, is equal to \dfrac{60}{80}=\dfrac{3}{4}
80
60
=
4
3
start fraction, 60, divided by, 80, end fraction, equals, start fraction, 3, divided by, 4, end fraction .
\qquad P(C)=\dfrac{3}{4}P(C)=
4
3
P, left parenthesis, C, right parenthesis, equals, start fraction, 3, divided by, 4, end fraction
Hint #22 / 5
Because there are 808080 total keys and 404040 open the teachers' lounge, we can say that the probability that a key opens the teachers' lounge, P(T)P(T)P, left parenthesis, T, right parenthesis, is equal to \dfrac{40}{80}=\dfrac{1}{2}
80
40
=
2
1
start fraction, 40, divided by, 80, end fraction, equals, start fraction, 1, divided by, 2, end fraction .
\qquad P(T)=\dfrac{1}{2}P(T)=
2
1
P, left parenthesis, T, right parenthesis, equals, start fraction, 1, divided by, 2, end fraction
Hint #33 / 5
Because there are 808080 total keys and 303030 open a classroom and the teachers' lounge, we can say that the probability that a key opens a classroom and the teachers lounge, P(C \text{ and } T)P(C and T)P, left parenthesis, C, start text, space, a, n, d, space, end text, T, right parenthesis, is equal to \dfrac{30}{80}=\dfrac{3}{8}
80
30
=
8
3
start fraction, 30, divided by, 80, end fraction, equals, start fraction, 3, divided by, 8, end fraction .
\qquad P(C\text{ and }T)=\dfrac{3}{8}P(C and T)=
8
3
P, left parenthesis, C, start text, space, a, n, d, space, end text, T, right parenthesis, equals, start fraction, 3, divided by, 8, end fraction
Hint #44 / 5
If we add the 606060 keys that open a classroom to the 404040 keys that open the teachers' lounge, we get a total of 100100100 keys. But this isn't quite right, because there are 303030 keys that open both types of room. We've counted them twice to get to 100100100, so we need to subtract 303030 to get the correct number of keys that open a classroom or the teachers' lounge.
\qquad \begin{aligned}P(C \text{ or } T) &= P(C) + P(T) - P(C\text{ and }T)\\ \\ &= \dfrac{60}{80}~~ + ~~\dfrac{40}{80} -~~~ \dfrac{30}{80}\\ \\ &=\dfrac{70}{80}\\ \\ &=\dfrac{7}{8}\end{aligned}
P(C or T)
=P(C)+P(T)−P(C and T)
=
80
60
+
80
40
−
80
30
=
80
70
=
8
7
This follows the Addition Rule of Probability, which states that given two events AAA and BBB,
P(A\text{ or }B) = P(A) + P(B) - P(A\text{ and }B)P(A or B)=P(A)+P(B)−P(A and B)P, left parenthesis, A, start text, space, o, r, space, end text, B, right parenthesis, equals, P, left parenthesis, A, right parenthesis, plus, P, left parenthesis, B, right parenthesis, minus, P, left parenthesis, A, start text, space, a, n, d, space, end text, B, right parenthesis.
Hint #55 / 5
The final answers are:
P(C)=\dfrac{3}{4}P(C)=
4
3
P, left parenthesis, C, right parenthesis, equals, start fraction, 3, divided by, 4, end fraction
P(T)=\dfrac{1}{2}P(T)=
2
1
P, left parenthesis, T, right parenthesis, equals, start fraction, 1, divided by, 2, end fraction
P(C\text{ and }T)=\dfrac{3}{8}P(C and T)=
8
3
P, left parenthesis, C, start text, space, a, n, d, space, end text, T, right parenthesis, equals, start fraction, 3, divided by, 8, end fraction
P(C\text{ or }T)=\dfrac{7}{8}P(C or T)=
8
7
Step-by-step explanation:
