A game is played, where a die is tossed and a marble selected from a bag. Bag M contains 3 red marbles (R) and 2 green marbles (G). Bag N contains 2 red marbles and 8 green marbles. A fair six-sided die is tossed. If a 3 or 5 appears on the die, bag M is selected (M). If any other number appears, bag N is selected (N). A single marble is then drawn at random from the selected bag. Question > Find the probability that a green marble is drawn from either bag.

2/3

Step-by-step explanation:

A die has 6 sides.  If a 3 or 5 is rolled, then bag M is selected.  So the probability that bag M is selected is 2/6 = 1/3.

Likewise, if a number other than 3 or 5 is rolled, then bag N is selected.  So the probability that bag N is selected is 4/6 = 2/3.

If bag M is selected, the probability the marble is green is 2/5.  The probability that M is selected, AND the marble is green, is therefore:

(1/3) (2/5) = 2/15

If bag N is selected, the probability that the marble is green is 8/10 = 4/5.  The probability that N is selected, AND the marble is green, is therefore:

(2/3) (4/5) = 8/15

The total probability that a green marble is selected is therefore:

P = 2/15 + 8/15

P = 10/15

P = 2/3

Bag M: 3 red and 2 green = 5 total, so green = 2/5

Bag N: 2 red and 8 green = 10 total, so green = 8/10 = 4/5

Dice roll: chance of getting Bag M is 2/6  ( 2 numbers out of 6 total numbers on a die), reduces to 1/3

Chance of getting Bag N is 4/6 = 2/3

The probability of picking green = Probability of Bag M x Probability of green in bag M + probability of bag N x probability of green in bag N:

2/5 * 1/3 + 4/5 * 2/3 = 2/15 + 8/15 = 10/15 = 2/3

The probability = 2/3

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the solution is h=10 because you add 9 to each side of the equation.