Question

Three balls are drawn one by one without replacement from a bag containing 5 white and 4 green .Find the probability distribution of number of green balls

Answer 1

Given:

5 white and 4 green balls.

To find:

probability distribution of number of green balls.

Solution:

1) Probability is the possibility of any event to be happened.

2) Let X be the random variable for the number of the green balls. X can take the values 0, 1, 2, 3.

3) Let A be the probability of getting first green ball and B be the probability of not getting first green ball.

∴ A = 4/9, B = 5/9

4) Probability distribution for the number of green balls

• P(X = 0) = BBB =  5 /9 ×4/8 ×3/7 = 60/504

• P(X= 1) = ABB + BAB + BBA

           = 4/9×5/8×4/7 + 5/9×4/8×4/7 + 5/9×4/8×4/7

           = 80/504 + 80/504 + 80/504

           = 240/504

• P (X=2) = AAB + ABA + BAA

            = 4/9×3/8×5/7 + 4/9×5/8×3/7 + 5/9×4/8×3/7

            = 60/504 + 60/504 + 60/504

            = 180/504

• P(X=3) = AAA = 4/9×3/8×2/7 = 24/504

Answer 2

Concept:

Probability is a way to gauge how likely something is to happen. Many things are difficult to forecast with absolute confidence. Using it, we can only make predictions about how probable an event is to happen, or its chance of happening. Probability can range from 0 to 1, with 0 denoting an impossibility and 1 denoting a certainty. For pupils in Class 10, probability is a crucial subject because it teaches all the fundamental ideas of the subject. A sample space has an overall probability of 1 for all events.

Given Information:

It is given that there are $5$ white balls and $4$ green balls in a bag.

Total number of balls in a bag = $9$ balls

To find:

If balls are drawn one by one without replacement from a bag, then find the probability distribution of number of green balls.

Solution:

The balls are drawn one by one without replacement from a bag three times.

Let X be the probability of drawing a green ball and Y be the probability of drawing the white ball.

P(X)= $\frac{4}{9}$

P(Y)= $\frac{5}{9}$

1- Probability when there are no green ball drawn three times:

   P(X=0) = $\frac{5}{9}* \frac{4}{8}*\frac{3}{7}$

               =$\frac{60}{504}$

2- Probability when there are one green ball is drawn when balls are drawn three times:

    P(X=1)= XYY + YXY + YYX

              = $\frac{4}{9}*\frac{5}{8}*\frac{4}{7} +\frac{5}{9}*\frac{4}{8}*\frac{4}{7} + \frac{5}{9}*\frac{4}{8}*\frac{4}{7}$

              =$\frac{240}{504}$

3- Probability when there are two green ball is drawn when balls are drawn three times:

    P(X=2)= XXY + XYX + YXX

               = $\frac{4}{9}*\frac{3}{8} *\frac{5}{7} + \frac{4}{9}*\frac{5}{8} *\frac{3}{7} +\frac{5}{9}*\frac{4}{8} *\frac{3}{7}$

               =$\frac{180}{504}$

4- Probability when there are three green ball is drawn when balls are drawn three times:

    P(X=3)= $\frac{4}{9}*\frac{3}{8}*\frac{2}{7}$

               =$\frac{24}{504}$

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