Respuesta :
Answer:
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
