Suppose a large jar contains eight red balls, six yellow balls, and six blue balls. Two balls are to be selected at random
from the jar, and the first ball selected will not be placed back into the jar.
(a) What is the probability that the first ball will be red and the second yellow?
(b) What is the probability that neither will be red?
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Answer:
Number of green balls = 3
Total number of balls = 6 + 3 + 5 +7 = 21
Probability of picking a green ball once = 3/21=1/7
Probability of an event happening twice when theevents are independant is simply their probability squared so:
Probability of picking 2 green balls = (1/7)^2 = 1/49
Explanation:
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