Question

One day, Karen bought 24 random cans of soup from a grocery store. Suppose that 7% of cans sold at that particular grocery store are dented, and assume the store's inventory is large enough that no individual customer's purchase affects the dent rate for the remaining cans. What is the probability that Karen has bought at least one dented can? Express your answer as a percentage precise to one decimal place.

Answer 1

Answer:

The probability that Karen has bought at least one dented can is 82.5%

Step-by-step explanation:

To know the probability of Karen buying at least one dented can, it's easier to calculate the probability of her not buying any dented can, and we know that:

$P_{\mbox{buying 24 cans}} =P_{\mbox{buying 24 non-dented cans}}+P_{\mbox{at least one dented can}}=1$

The probability of a can not being dented is (in the same principle as above, 100%(all cans)-7%(dented cans)=93%(non-dented cans) 0.93.

As the probability of a can being dented or not is independent from each other, we multiply the probabilities:

$P_{\mbox{buying 24 non-dented cans}}=0.93^{24}=0.175$

Now, we calculate the probability of at least one dented can, expressed as a percentage:

$P_{\mbox{at least one dented can}}= (1-P_{\mbox{buying 24 non-dented cans}})*100\\\\P_{\mbox{at least one dented can}}=(1-0.175)*100=82.5\%$