asked 200k views
0 votes
A goblet contains 2 red marbles, 6 green marbles, and 4 blue marbles.

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

asked
User Mweber
by
7.8k points

2 Answers

3 votes

Answer:

5/22

Explanation:

answered
User Torbatamas
by
8.3k points
5 votes

Answer:

P(first marble will be green and the second will be green as well)= 5/22

Explanation:

A goblet contains 2 red marbles, 6 green marbles, and 4 blue marbles.

Total number of marbles= 2+6+4

=12

P(first marble is green)=no. of green marbles/total number of marbles

= 6/12

= 1/2

P(second marble is green and we didn't put the first marble back)

=no. of green marbles left/total marbles left

= 5/11

Hence, P(first marble will be green and the second will be green as well)

=
(1)/(2)* (5)/(11)

= 5/22

answered
User Justin Skiles
by
8.1k points
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