A bag contains 2 red marbles, 3 green marbles, and 4 blue marbles.
If we choose a marble, then another marble without putting the first one back in the bag, what is the probability that the first marble will be green and the second will be red?
It's on Khan Academy of Pre-Calculus. It needs to be 100% correct.
Answers
Answer:
1/14
Step-by-step explanation:
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The probability will be 1/12
Step-by-step explanation:
Given:
A bag contains 2 red marbles, 3 green marbles, & 4 blue marbles
Total number of marbles = 9
To find:
the probability that the first marble will be green and the second will be red when a marble is chosen without putting the first one in the bag
Solution:
The total marbles will be 9
Probability of getting the green marble at first= number of green marbles
/total number of marbles
Probability of getting one green marble= 3/9 = 1/3
Now, we have 8 marbles left after one green marble is drawn
Probability of getting red marble at second= number of red marbles
/total number of marbles
Probability of getting red marble at second draw= 2/8 = 1/4
Now, the probability that the first marble will be green and the second one will be red= (1/3)x(1/4)
The probability that the first marble will be green and the second one will be red= 1/12
Thus, the probability that the first marble will be green and the second will be red is 1/12
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