A bag contains 3 red and 5 black balls and second bag contains 6 red and 4 black balls. A ball is drawn from each bag. Find the probability that one is red and the other is black.

To find the probability that one ball is red and the other is black when drawing from two bags, we first need to determine the total number of balls in each bag and the possible outcomes.

Details of the Bags

• First Bag: 3 red balls, 5 black balls (Total: 8 balls)
• Second Bag: 6 red balls, 4 black balls (Total: 10 balls)

Possible Outcomes

We can have two scenarios for drawing one red and one black ball:

• Red from the first bag and black from the second bag
• Black from the first bag and red from the second bag

Calculating Probabilities

Let's calculate the probabilities for each scenario:

Scenario 1: Red from Bag 1 and Black from Bag 2

The probability of drawing a red ball from the first bag is:

P(Red from Bag 1) = 3/8

The probability of drawing a black ball from the second bag is:

P(Black from Bag 2) = 4/10 = 2/5

The combined probability for this scenario is:

P(Red from Bag 1 and Black from Bag 2) = (3/8) * (2/5) = 6/40 = 3/20

Scenario 2: Black from Bag 1 and Red from Bag 2

The probability of drawing a black ball from the first bag is:

P(Black from Bag 1) = 5/8

The probability of drawing a red ball from the second bag is:

P(Red from Bag 2) = 6/10 = 3/5

The combined probability for this scenario is:

P(Black from Bag 1 and Red from Bag 2) = (5/8) * (3/5) = 15/40 = 3/8

Total Probability

Now, we add the probabilities of both scenarios to find the total probability of drawing one red and one black ball:

Total Probability = (3/20) + (3/8)

Finding a Common Denominator

The least common multiple of 20 and 8 is 40. Converting both fractions:

• 3/20 = 6/40
• 3/8 = 15/40

Now, adding these:

Total Probability = 6/40 + 15/40 = 21/40

Final Result

The probability that one ball is red and the other is black is 21/40.