a bag contains 4 red and 7 green balls. if three balls are drawn from the bag replaced and once again three balls are drawn from the bag what is the probability of obtaining 3 red balls on the first drawing and 3 green balls on the second drawing?
Answers
Therefore the probability of obtaining 3 red balls on the first draw and 3 green balls on the second draw is 28/121.
Given:
Number of red balls in the bag = 4 balls
Number of green balls in the bag = 7 balls
3 balls are drawn from the bag and replaced and again drawn a second time.
To Find:
Probability of obtaining 3 red balls on the first draw and 3 green balls on the second draw.
Solution:
The given question can be solved as shown below.
Given that,
Number of red balls in the bag = 4 balls
Number of green balls in the bag = 7 balls
On the first draw:
Probability of getting 3 red balls = ( number of favorable outcomes )/(number of total outcomes)
⇒ Probability of getting 3 red balls = ⁴C₃ / ¹¹C₃
⇒ Probability of getting 3 red balls = 4/11
On the second draw:
Probability of getting 3green balls = ( number of favorable outcomes )/(number of total outcomes)
⇒ Probability of getting 3 red balls = ⁷C₃ / ¹¹C₃
⇒ Probability of getting 3 red balls = 7/11
Now the probability of getting 3 red balls on the first draw and replaced again and getting 3 green balls on the second draw = (4/11) × (7/11) = 28/121
Therefore the probability of obtaining 3 red balls on the first draw and 3 green balls on the second draw is 28/121.
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