Question

a bag contains 6 blue balls, 3 green balls and 3 yellow balls. what is the probability of choosing one blue ball and then a yellow ball with replacement? responses 34 three fourths 322 begin fraction numerator 3 over denominator 22 end fraction 18 one eighth 116

Answer 1

P = 1/8 (0.125)

Explanation:

First selection: P = 6/12 = 1/2

Second selection: P = 3/12 = 1/ 4

Probability of selecting a blue and yellow ball with replacement:

`P=((1)/(2) )((1)/(4) )=((1)(1))/((2)(4)) =(1)/(8)`

Decimal: P = 0.125

Hope this helps

Answer 2

Answer: The probability of choosing one blue ball and then a yellow ball with replacement is 1/4.

Explanation:

When choosing a ball with replacement, it means that after each selection, the ball is placed back into the bag. Since there are 6 blue balls out of a total of 12 balls in the bag, the probability of choosing a blue ball on the first draw is 6/12, which simplifies to 1/2. After replacing the blue ball, there are still 12 balls in the bag, with 3 of them being yellow. Therefore, the probability of choosing a yellow ball on the second draw is 3/12, which simplifies to 1/4. To find the overall probability, we multiply the probabilities of each event together, resulting in (1/2) * (1/4) = 1/4.