Question

A box contains 5 white and 7 black balls. two successive drawn of 3 balls are made (i) with replacement (ii) without replacement . the probability that the first draw would produce white balls and the second draw would produce black balls are respectively

Answer 1

Two successive drawn of 3 balls are made:
TOTAL = 12: 5W, 7B 
1st draw(a) with replacement 
Draw 3
P(3-W) = (5/12)(5/12)(5/12) = 125/1728
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2nd draw(b)without replacement.
Draw 3
P(3-B) = (7/12)(6/11)(5/10) = 210/1320 = 21/132
P(both happening) = (125/1728)(21/132) 

Answer 2

Answer:

Probability of both happening = $\frac{125}{1728}\times\frac{21}{132}=\frac{2625}{228096}$

Step-by-step explanation:

Given : A box contains 5 white and 7 black balls. Two successive drawn of 3 balls are made (i) with replacement (ii) without replacement .

To find : The probability that the first draw would produce white balls and the second draw would produce black balls are respectively.

Solution :

$\text{Probability}=\frac{\text{Favorable outcome}}{\text{Total number of outcome}}$

A box contains 5 white and 7 black balls = 12 balls

i) On a first drawn, (with replacement)

Total number of outcome = 12

First draw would produce white balls

Probability of first drawn is  

$\text{Probability}=\frac{5}{12}\times\frac{5}{12}\times\frac{5}{12}=\frac{125}{1728}$

ii) On a second drawn, (without replacement)

Total number of outcome = 12

The second draw would produce black balls

Probability of second drawn is  

$\text{Probability}=\frac{7}{12}\times\frac{6}{11}\times\frac{5}{10}=\frac{210}{1320}=\frac{21}{132}$

→ Probability of both happening = $\frac{125}{1728}\times\frac{21}{132}=\frac{2625}{228096}$