A batch consists of 12 defective coils and 88 good ones. find the probability of getting two good coils when two coils are randomly selected if the first selection is replaced before the second is made.
Help me..
since it would be replaced would the answer be 0.7744 ?
To find the probability of getting two good coils when two coils are randomly selected, with the first selection replaced before the second is made, we can use the concept of probability.
First, let's determine the total number of coils in the batch. Since the batch consists of 12 defective coils and 88 good ones, the total number of coils is 12 + 88 = 100.
To find the probability of getting two good coils, we can calculate the probability of getting a good coil on the first selection and multiply it by the probability of getting a good coil on the second selection. Since the first selection is replaced before the second is made, the probability remains the same for both selections.
The probability of selecting a good coil on any single draw is given by the ratio of the number of good coils to the total number of coils. In this case, the probability is 88/100 = 0.88.
Therefore, the probability of getting two good coils can be calculated as follows:
Probability = (0.88) * (0.88) = 0.7744
So, yes, your answer of 0.7744 is correct.