A flashlight has 8 batteries, 2 of which are defective. If you select 2 batteries, find the probability that
You select at least 1 good battery
Probability of getting either one or two good batteries.
Either-or probabilities are found by adding the individual probabilities.
If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
P(1 good) = 6/8 * 2/7 = ?
P(2 good) = 6/8 * 5/7 = ?
0.75
To find the probability of selecting at least 1 good battery, we need to consider two cases: selecting 1 good battery and selecting 2 good batteries. We will then calculate the probability for each case and add them together.
Case 1: Selecting 1 good battery:
To calculate the probability of selecting 1 good battery, we will use the concept of combinations.
There are a total of 8 batteries, and we need to select 2 batteries. Since 2 batteries are defective, there are 6 good batteries. The number of ways to select 1 good battery out of 6 is given by the combination formula: C(6, 1) = 6!/[(6-1)! * 1!] = 6.
The total number of ways to select any 2 batteries out of 8 is given by the total number of combinations: C(8, 2) = 8!/[(8-2)! * 2!] = 28.
Therefore, the probability of selecting 1 good battery is 6/28.
Case 2: Selecting 2 good batteries:
To calculate the probability of selecting 2 good batteries, we will again use the combination formula.
There are 6 good batteries out of 8. The number of ways to select 2 good batteries out of 6 is given by: C(6, 2) = 6!/[(6-2)! * 2!] = 15.
The total number of ways to select any 2 batteries out of 8 is 28, as calculated before.
Therefore, the probability of selecting 2 good batteries is 15/28.
Now, to find the probability of selecting at least 1 good battery, we add the probabilities from each case:
P(at least 1 good battery) = P(1 good battery) + P(2 good batteries)
= 6/28 + 15/28
= 21/28
= 0.75
So, the probability of selecting at least 1 good battery is 0.75 or 75%.