There are 1010 black balls and 77 red balls in an urn. If 44 balls are drawn without replacement, what is the probability that no more than 11 black ball is drawn? Express your answer as a fraction or a decimal number rounded to four decimal places

P (no or one Black ball) = 0.1618

Step-by-step explanation:

* The errors in the question are underlined*

Given:-

An urn contains the following balls:

- Number of black balls, B = 10

- Number of red balls, R = 7

Find:-

If 4 balls are drawn without replacement, what is the probability that no more than 1 black ball is drawn?

Solution:-

- The urn contains Black and Red colored balls. The total number of balls (n) in the urn are:

n = B + R

n = 10 + 7

n = 17 balls

- We are to draw 4 balls from the urn. This question implies for a selection of balls from urn hints the application of combinations.

- The total number of possible combinations for randomly drawing r = 4 balls from an urn containing n = 17 balls is:

All combinations = 17C4

= 2380 (without any restriction)

- However, the restriction imposed is to select such 4 balls that have no more than 1 black ball. This can be broken down into cases where there is either no or 1 black ball out of selected 4. The possible number of combination are:

no or one Black ball = No black ball + 1 black ball

= (Choose 4 Red) + (1 Black & 3 Red)

= 7C4 + (10C1 * 7C3)

= 35 + (10*35)

= 35 + 350

= 385 (with restriction)

- The probability for our restricted case is as follows:

P (no or one Black ball) = with restriction / without restriction

= 385 / 2380

= 0.1618