McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
6. Probabilities of Mutually Exclusive Events
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Exercise 14 Page 929

If A and B are events that have outcomes in common, then P(AorB) = P(A)+P(B)-P(AandB).

14/25 or 56 %

Practice makes perfect

Events are mutually exclusive when they cannot occur at the same time. Or, they are considered overlapping when they have outcomes in common. The following rules apply for the probability of these types of compound events.

Mutually Exclusive Events Overlapping Events
P(AorB)= P(A)+P(B) P(AorB)= P(A)+P(B)-P(AandB)
Let A be player is 14 and B be player plays basketball. Since some students are both 14 and play basketball, these events are not mutually exclusive. Now, we can find the number of players aged 14 and the number of basketball players aged 14 in the given table.
Graceland Sports Complex
Age Soccer Baseball Basketball Total
14 28 36 42 28+36+ 42= 106
15 30 26 33 30+26+33=89
16 35 41 29 35+41+29=105

There are 106 players aged 14. Out of these players, 42 play basketball. Next, we can calculate the number of basketball players and the number of all players in the table.

Graceland Sports Complex
Age Soccer Baseball Basketball Total
14 28 36 42 106
15 30 26 33 89
16 35 41 29 105
Total 28+30+35=93 36+26+41=103 42+33+29= 104 106+89+105= 300
Now, we know that there are 106 players aged 14, 104 basketball players, 42 basketball players aged 14, and 300 players in total. Finally, we can calculate P(A), P(B), and P(AandB). P(A)&=106/300 l← ← lAged14 Total players P(B)&=104/300 l← ← lBasketball Total players P(AandB)&=42/300 l ← ← lBasketball and aged14 Total players With this information we want to find the value of P(AorB).
P(AorB)=P(A)+ P(B)-P(AandB)
P(Aor)= 106/300+ 104/300- 42/300
Simplify right-hand side
P(AorB)=168/300
P(AorB)=14/25
Convert to percent
P(AorB)=0.56
P(AorB)=56 %
The probability that a player is 14 or plays basketball is equal to 1425, which can be also written as 56 %.