Question

question 6 1 pts a jar contains black and white marbles. two marbles are chosen without replacement. the probability of selecting a black marble and then a white marble is 34%, and the probability of selecting a black marble on the first draw is 47%. what is the probability of selecting a white marble on the second draw, given that the first marble drawn was black? (write your answer as a percentage rounded to 1 decimal)

Explanation:

Step1: Recall conditional - probability formula

The formula for conditional probability is $P(B|A)=\frac{P(A\cap B)}{P(A)}$. Let $A$ be the event of drawing a black marble first and $B$ be the event of drawing a white marble second. We know that $P(A\cap B) = 0.34$ (probability of black then white) and $P(A)=0.47$ (probability of black on first draw).

Step2: Substitute values into formula

$P(B|A)=\frac{0.34}{0.47}$.

Step3: Calculate the result

$P(B|A)\approx 0.7234$.

Step4: Convert to percentage and round

$0.7234\times100 = 72.34\%$, rounded to one - decimal place is $72.3\%$.

Answer:

$72.3\%$