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
prob(both defective) = (2/8)(1/7) = 1/28
prob(at least one good) = 1 - 1/28 = 27/28
8c2= 28
Picking one (2c1)(6c1)=12
prob= 12/28=.43
To find the probability of selecting at least 1 good battery, we can use the concept of complementary probability.
First, let's determine the total number of ways to select 2 batteries out of the 8 batteries. This can be calculated using the combination formula:
C(n, r) = n! / (r!(n-r)!)
Where n is the total number of batteries and r is the number of batteries we want to select. In this case, n = 8 and r = 2, so we can substitute these values into the formula:
C(8, 2) = 8! / (2!(8-2)!) = 8! / (2!6!)
Simplifying further:
C(8, 2) = (8 * 7 * 6!) / (2! * 6!)
The 6! terms cancel out, leaving us with:
C(8, 2) = (8 * 7) / (2 * 1) = 28
Therefore, there are 28 different combinations of selecting 2 batteries from the 8 available.
Now let's consider the case where no good batteries are selected. Since there are 2 defective batteries, the number of ways to select 2 defective batteries can be calculated using the combination formula again with n = 2 and r = 2:
C(2, 2) = 2! / (2!(2-2)!) = 2! / (2! * 0!) = 1
So there is 1 way to select 2 defective batteries.
Finally, to find the probability of selecting at least 1 good battery, we can subtract the probability of selecting only defective batteries from 1:
P(select at least 1 good battery) = 1 - P(select 2 defective batteries)
P(select 2 defective batteries) = 1 / 28
Therefore,
P(select at least 1 good battery) = 1 - (1 / 28)
= (28 / 28) - (1 / 28)
= 27 / 28
Hence, the probability of selecting at least 1 good battery from the flashlight is 27/28.