Multiple choice

A bag contains 5 red and 3 green balls. Another bag contains 4 red and 6 green balls. If one ball is drawn from each bag. Find the probability that one ball is red and one is green.

  1. $\dfrac{23}{40}$
  2. $\dfrac{21}{40}$
  3. $\dfrac{19}{40}$
  4. $\dfrac{17}{40}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

There are two scenarios: (Red from Bag 1 AND Green from Bag 2) OR (Green from Bag 1 AND Red from Bag 2). Probabilities are (5/8 * 6/10) + (3/8 * 4/10) = 30/80 + 12/80 = 42/80 = 21/40.

AI explanation

The probability of drawing a red ball from the first bag is 5/8 and a green ball from the second bag is 6/10, giving (5/8) multiplied by (6/10). The probability of drawing a green ball from the first bag is 3/8 and a red ball from the second bag is 4/10, giving (3/8) multiplied by (4/10). Adding these two mutually exclusive probabilities yields 30/80 plus 12/80, which equals 42/80 or 21/40. The probability is 21/40.

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