In: Statistics and Probability
Harris and Mya each have a jar that contains 6 marbles. In each jar, d of the marbles are blue and the remaining marbles are green.
Harris will take out a marble from his jar at random and then replace it. He will then take out another marble from his jar at random and replaces it.
Mya will take out two marbles from her jar at random without replacement.
The probability that Mya observes two blue marbles is 80% of the probability that Harris observes two blue marbles.
(a) How many green marbles and blue marbles are in each jar?
(b) Harris then will take out 3 marbles from his jar at random without replacement. What is the probability that his sample contains at least 2 green marbles?
Solution
Part (a)
Since Harris’ draw is one after another with replacement, probability his draw of two marbles will have two blue marbles
= (d/6)(d/6)
= d2/36 ..................................................................................................................................................................................... (1)
Since Mya’s draw is without replacement, probability her draw of two marbles will have two blue marbles
= {d(d - 1)}/(6 x 5)
= {d(d - 1)}/30........................................................................................................................................................................... (2)
By the given condition, ‘The probability that Mya observes two blue marbles is 80% of the probability that Harris observes two blue marbles.’ along with (1) and (2)
=> {d(d - 1)}/30 = 0.8d2/36
Or, d = 3
Thus, number of green marbles and blue marbles in each jar is 3 and 3. Answer 1
Part (b)
When Harris takes out 3 marbles from his jar at random without replacement,
probability that his sample contains 2 green marbles, i.e., 2 green and 1 blue = {(3C2)(3C1)}/6C3 = 0.45 and
probability that his sample contains all 3 green marbles = 3C3/6C3 = 0.05
Hence, probability that sample of Harris contains at least 2 green marbles
= 0.45 + 0.05
= 0.5 Answer 2
DONE