A box contains 22 red marbles, 55 green marbles and 55 blue marbles. If we choose a marble, then another marble without putti

Question

3 years ago

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A box contains 22 red marbles, 55 green marbles and 55 blue marbles. If we choose a marble, then another marble without putti

ng the first one back in the box, what is the probability that the first marble will be green and the second will be blue?

Mathematics

1 answer:

PSYCHO15rus [73] · Answer 1 · 2022-03-13T23:16:55+03:00

Answer:

3025/17292

Step-by-step explanation:

A box contains 22 red marbles, 55 green marbles and 55 blue marbles.

The total number of marbles in the box = 22 red marbles + 55 green marbles + 55 blue marbles

= 132 marbles

If we choose a marble, then another marble without putting the first one back in the box, what is the probability that the first marble will be green and the second will be blue?

This means we are dealing with probability without replacement.

Hence:

P(Green) and P(Blue)

= P(Green) × P(Blue)

= 55/132 × 55/131

= 3025/17292

Therefore, the probability that the first marble will be green and the second will be blue is 3025/17292

A box contains 22 red marbles, 55 green marbles and 55 blue marbles. ​ ​If we choose a marble, then another marble without putti

A box contains 22 red marbles, 55 green marbles and 55 blue marbles. If we choose a marble, then another marble without putti