A bag contains 6 red balls, 4 green balls, and 3 blue balls. If we choose a ball, then another ball without putting the first one back in the bag, what is the probability that the first ball will be green and the second will be red?
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Answered by
3
Answer:
Probability of first ball Will be green is 4/13.
Probability of second will be red is 6/12 that is equal to1/2
May it help you
Answered by
2
p(first ball)=4/13
p(second ball)=6/12=1/2
Step-by-step explanation:
p(first ball)=no.of green balls/total no.of balls
=4/13
one ball is removed then total no.of balls=12
p(second ball)=no.of red balls/total no.of balls
=6/12
=1/2
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