Question

a) A bag contains 10 white and 6 black balls. Four balls are drawn
out one by one without replacement. What is the probability of
getting balls of alternatively different colours ?
​

Answer 1

Answer:

They are two ways of picking balls of alternating colours. You could draw WBWB or BWBW.

The probability of picking WBWB is (10/16)x(6/15)x(9/14)x(5/13) and of picking BWBW is (6/16)x(10/15)x(5/14)x(9/13).

Both probabilities are equal to 2700/43680 or 45/728.

Therefore 45/728+45/728 = 45/364.

This is equal to about a 12.36% chance of picking balls of alternating colours.

Answer 2

Answer:

Probability of getting alternatively different color = $\frac{45}{364}$.

Step-by-step explanation:

Given that A bag contains 10 white and 6 black balls. Four balls are drawn out one by one without replacement.  

Probability (Favorable outcome) = $\frac{Number Of Favorable Outcome}{Total Outcome}$

The possible sequence of getting balls of alternatively different colors are;

(i) White, Black, White, Black

(ii) Black, White, Black, White

(i) White, Black, White, Black

First ball is white, with probability $\frac{10}{16}$ (10 white balls, 16 total balls left)

Second ball is Black, with probability $\frac{6}{15}$ (6 black balls, 15(16-1) total balls left)

Third ball is white, with probability $\frac{9}{14}$ (9(10-1) white balls, 14(16-2) total balls left)

Fourth ball is black, with probability $\frac{5}{13}$ (5(6-1) black balls, 13(16-3) total balls left)

Probability of getting the sequence WBWB = $\frac{10}{16}*\frac{6}{15}*\frac{9}{14} *\frac{5}{13} = \frac{2700}{43,680}$

(ii) Black, White, Black, White

First ball is Black, with probability $\frac{6}{16}$ (6 black balls,16 total balls left)

Second ball is white, with probability $\frac{10}{15}$ (10 white balls, 15(16-1) total balls left)

Third ball is black, with probability $\frac{5}{14}$ (5(6-1) black balls, 14(16-2) total balls left)

Fourth ball is white, with probability $\frac{9}{13}$ (9(10-1) white balls, 13(16-3) total balls left)

Probability of getting the sequence BWBW = $\frac{6}{16}* \frac{10}{15}*\frac{5}{14}* \frac{9}{13} = \frac{2700}{43,680}$

Total possible outcomes = $\frac{2700}{43,680} + \frac{2700}{43,680} = \frac{5400}{43,680} = \frac{45}{364}$.

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