Question

question 7
1 pts
in a bag, there are two types of candy bars: snickers and twix. two items are chosen at random from the bag, without replacement. the probability of selecting a snickers on the first draw is 71%. the probability of selecting a snickers and then a twix is 47%. what is the probability selecting a twix on the second draw, given that the first was a snickers? (write your answer as a percentage rounded to 1 decimal)

Explanation:

Step1: Recall conditional - probability formula

Let $A$ be the event of selecting a Snickers on the first draw and $B$ be the event of selecting a Twix on the second draw. The formula for conditional probability is $P(B|A)=\frac{P(A\cap B)}{P(A)}$.

Step2: Identify given probabilities

We are given that $P(A) = 0.71$ (probability of selecting a Snickers on the first draw) and $P(A\cap B)=0.47$ (probability of selecting a Snickers and then a Twix).

Step3: Calculate the conditional probability

Substitute the given values into the formula: $P(B|A)=\frac{0.47}{0.71}\approx0.662$.

Step4: Convert to percentage and round

Multiply by 100 to get a percentage: $0.662\times100 = 66.2\%$.

Answer:

$66.2\%$