# Another Problem- Can't figure it out.

• Mar 31st 2009, 01:25 AM
Curious_eager
Another Problem- Can't figure it out.
A bag contains 5 Black and 3 red balls.A ball is taken out of a bag and is not returned to it.If the process is treated 3 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.
• Mar 31st 2009, 04:39 AM
Curious_eager
Thought process -

I think the three balls can be drawn in 4 ways

1) All the three balls are black = 5/8 .4/7 .3/6 then the probability of next black ball can be drawn is 2/5.
2) 2 black and 1 red = 5/8 . 4/7 . 3/6 then the probability of next black ball can be drawn is 3/5.
3) 1 black and 2 red = 5/8 . 3/7 . 2/6 then the probability of next black ball can be drawn is 4/5.
4) All red = 3/8 . 2/7 . 1/6 then the probability of next black ball can be drawn is 1.

Probability = (option1 + option 2 + option3 + option 4)/chosing one out of four

= (2/5 + 3/5 + 4/5 + 1)/4C1 = 14/20 = 0.7

Please let me know if the way I am thinking is right.

Anshu.
• Mar 31st 2009, 05:17 AM
mr fantastic
Quote:

Originally Posted by Curious_eager
A bag contains 5 Black and 3 red balls.A ball is taken out of a bag and is not returned to it.If the process is treated 3 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.

One way of doing this would be to draw a tree diagram.
• Mar 31st 2009, 06:21 AM
Curious_eager
Thanks Mr Fantastic, Just want to know If the approach I followed is right or wrong.
• Mar 31st 2009, 08:19 AM
Plato
Quote:

Originally Posted by Curious_eager
1) All the three balls are black = 5/8 .4/7 .3/6 then the probability of next black ball can be drawn is 2/5.
2) 2 black and 1 red = 5/8 . 4/7 . 3/6 then the probability of next black ball can be drawn is 3/5.
3) 1 black and 2 red = 5/8 . 3/7 . 2/6 then the probability of next black ball can be drawn is 4/5.
4) All red = 3/8 . 2/7 . 1/6 then the probability of next black ball can be drawn is 1.

You have correctly identified the four possibilities.
But what you have not done correctly is to calculate the probabilities of each option.
For example, #2 should be: $\displaystyle {3 \choose 1}\frac{5\cdot 4\cdot 3}{8\cdot 7\cdot 6} \frac {3}{5}$.

If done correctly by adding we get $\displaystyle 0.625$.
It is curious to note that is exactly the probability of getting a black on the first draw.