Big Ideas Math Geometry, 2014
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Big Ideas Math Geometry, 2014 View details
4. Probability of Disjoint and Overlapping Events
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Exercise 15 Page 698

Recall the formula for the probability of compound events. Find the probabilities from the formula, substitute them into the formula, and isolate P(AandB).

10%

Practice makes perfect
We are told that the botanist determined that in a group of 40 improperly growing trees 34 trees have a disease or are being damaged by insects, 18 trees have a disease, and 20 are being damaged by insects. We want to find the probability of randomly picking a tree which both has a disease and is being damaged by insects. To find it we will use the formula for the probability of compound events. P(AorB)=P(A)+P(B)-P(AandB) In our case event A will represent the tree having a disease and event B will represent the tree being damaged by insects. Since we know that 34 trees have a disease or are being damaged by insects, we can calculate P(A or B). To do so we will divide the number of trees that have a disease or are being damaged by insects, 34, by the total number of trees, 40.
P(AorB)=Disease or insects/Total number of trees
P(AorB)=34/40
Now we will repeat the same procedure for the probability of event A and the probability of event B. We know that 18 trees have a disease and 20 trees are being damaged by insects. To calculate probabilities of events A and B we need to divide those numbers by the total number of the trees, 40.
Probability
P(A)=Number of trees with the disease/Total number of trees P(B)=Number of trees damaged by insects/Total number of trees
P(A)=18/40 P(B)=20/40
Note that now we know every value from the formula for the probability of compound events except the value of P(AandB). Therefore, we can substitute the obtained values into this formula and isolate P(AandB). Let's do it!
P(AorB)=P(A)+P(B)-P(AandB)
34/40=18/40+20/40-P(AandB)
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Solve for P(AandB)
34/40=38/40-P(AandB)
34/40-38/40=- P(AandB)
-4/40=- P(AandB)
4/40=P(AandB)
P(AandB)=4/40
P(AandB)=0.1
The probability of randomly selecting a tree that both has a disease and is being damaged by insects is 0.1, or 10%.