Answer to Question #161512 in Statistics and Probability for Learner123

Question

There are three blue counters and six green counters in a bag. A counter is picked out of the bag, the colour noted and then replaced three times. The random variable X is the number of blue counters taken from the bag.

a) find the probability of distribution of X

b) find the probability of getting one or more blue counters

Answer 1

The number of ways to take 4 counters out of 9 is 9!/(4!5!)=126

The number of ways to take 4 green counters out of 6 is 6!/(4!2!)=15

The number of ways to take 3 green counters and 1 blue counter out of 3 is 6!/(3!3!)*3!/(1!2!)=60

The number of ways to take 2 green counters and 2 blue counters is 6!/(2!4!)*3!/(2!1!)=45

The number of ways to take 1 green counter and 3 blue counters is 6!/(1!5!)*3!/(3!1!)=6

The number of ways to take 4 blue counters is 0.

The random variable X is the number of blue counters taken from the bag.

Its possible values are 0, 1, 2, 3. Their probabilities are

P(X=0)=15/126=0.119

P(X=1)=60/126=0.4762

P(X=2)=45/126=0.3571

P(X=3)=6/126=0.0476

(b) $P(X\geq1)=1-P(X=0) = 1-0.119=0.881$