There are 6 red and 5 green balls in the urn, differing only in color. a) one ball is taken out at random.

| May 18, 2021 | education

There are 6 red and 5 green balls in the urn, differing only in color. a) one ball is taken out at random. Find the probability that it is green; b) Find the probability that one ball is red and the other green

There are 6 + 5 = 11 balls in the urn.
a) Probability that the removed ball is green:
P = 5/11 = 0.4545.
b) The probability that the red ball was taken out first, and then the green one:
P1 = 6/11 * 5/10 = 3/11.
If you first took out the red ball, then 10 balls remain, of which 5 are green.
You can first take out the green ball, and then the red one.
The probability of this event will be equal to:
P2 = 5/11 * 6/10 = 3/11.
The probability that the red and green balls will be taken out in any sequence will be equal to the sum of the probabilities P1 and P2, since these are inconsistent events.
P1 + P2 = 3/11 + 3/11 = 6/11 = 0.5454.
Answer: The probability of taking out a green ball is 0.4545, one green and one red is 0.5454.



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