Question

A bag contains 5 black and 3 red balls. a ball is taken out of the bag and is not returned to it . if this process is repeated three times then what is the probability of drawing a red ball in the 4th draw of a ball?

Answer 1

The probability of drawing a red ball in the 4th draw of a ball is 11/56.

Step-by-step explanation:

The chances of fourth ball being a red ball are :-

Condition one:-  If first three balls drawn out are black then chances of 4th ball being red are

$\frac{5}{8} \times \frac{4}{7} \times \frac{3}{6} \times \frac{3}{5} = \frac{3}{28}$

Condition two :-  If 3 balls drawn out are red  then the chances of the 4th ball being red one are 0

Condition three :-  If two red balls are drawn out and one black ball is drawn out then chances of 4th ball being red $\frac{3}{8}\times\frac{2}{7}\times\frac{5}{6}\times\frac{1}{5}=\frac{1}{56}$

Condition four :-  If one red Ball and 2 black balls are drawn out then chances of fourth ball being red are $\frac{3}{8}\times\frac{5}{7}\times\frac{4}{6}\times\frac{2}{5}=\frac{1}{14}$

Then chances of 4th ball being red are = $\frac{3}{28} + \frac{1}{56} + \frac{1}{14} = \frac{11}{56}$

 

To know more :-

A bag contains 5 black and 3 red balls. a ball is taken out of the bag and is not returned to it . if this process is repeated three times then what is the probability of drawing a black ball in the next draw of a ball.

https://brainly.in/question/5029530

A bag contains 5 black and 3 red balls. A ball is taken out of the bag and is not returned to it. If this process is repeated three times, then what is the probability of drawing a black ball in the next draw of a ball? (a) 0.7 (b) 0.625 (c) 0.1 (d) none of these

https://brainly.in/question/13382749

Answer 2

Given:

A bag contains 5 black and 3 red balls

a ball was taken out and not return it.

To find:

the probability of drawing a red ball in the 4th.

Solution:

The probability for getting all black ball:

$\Rightarrow \frac{5}{8} \times \frac{4}{7} \times \frac{3}{6}$

The next red ball which can be drawn is:

$\Rightarrow \frac{2}{5}$

The probability of getting 2 red and 1 black balls:

$\Rightarrow \frac{5}{8} \times \frac{4}{7} \times \frac{3}{6}$

The next red ball which can be drawn is:

$\Rightarrow \frac{3}{5}$

The probability of getting 1 black and 2 red balls:

$\Rightarrow \frac{5}{8} \times \frac{3}{7} \times \frac{2}{6}$

The next red ball which can be drawn is:

$\Rightarrow \frac{4}{5}$

The probability of getting only red ball:

$\Rightarrow \frac{3}{8} \times \frac{2}{7} \times \frac{1}{6}$

The one red ball which can be drawn is:

$\Rightarrow 1$

And the chances for every route are 1/4. This is why you want a chance:

$\Rightarrow \frac{1}{4} \times \frac{2}{5}+\frac{1}{4} \times \frac{3}{5} + \frac{1}{4} \times \frac{4}{5}+ \frac{1}{4} \times 1\\\\\Rightarrow \frac{7}{10} \\\\\Rightarrow 0.7$