# Thread: A rather hard probability question.

1. ## A rather hard probability question.

Hi there.

The question is:

A bag contains eight counters. Five of the counters are green, three are red. A counter is taken at random from the bag and its colour is noted. The counter is not replaced. A second counter is then taken from the bag. Calculate the probability that:

(i) the first counter is green and the second is red
(ii) the counters are of the same colour
I've done (i), it was pretty basic. But (ii) has me stumped. Anyone got any ideas?

Thanks in advance,

- Chris Down

2. Originally Posted by Chris Down
Hi there.

The question is:

I've done (i), it was pretty basic. But (ii) has me stumped. Anyone got any ideas?

Thanks in advance,

- Chris Down

I'm not really good with probability, so excuse me if these are not completely correct.

Five of the counters are green, three are red.
For 1)

We have 8 counters in total, the chances of drawing a green counter would be 5 out of 8, which is 62,5 %

When the 2nd counter is drawn, there are only 7 counters left.

Chances of drawing a red counter would then be 3 out of 7, which is a chance of 42,9%

The chances that the first one is green AND the second one is red is:

$\frac{5}{8} \times \frac{3}{7}$

$= \frac{15}{56}$

$= 26,8$ %

-------------------

For 2)

The chances of drawing two green counters are:

$\frac{5}{8} \times \frac{4}{7}$

$= \frac{5}{14}$

$= 35,7$ %

The chances of drawing two red counters are:

$\frac{3}{8} \times \frac{2}{7}$

$= \frac{3}{28}$

$= 10,7$ %

So the chances of drawing either a green pair, or a red pair is:

$\frac{5}{14} + \frac{3}{28}$

$= \frac{13}{28}$

$= 46,4$ %

3. Originally Posted by janvdl
The chances of drawing two green counters are:

$\frac{5}{8} \times \frac{4}{7}$

$= \frac{5}{14}$

$= 35,7$ %

The chances of drawing two red counters are:

$\frac{3}{8} \times \frac{2}{7}$

$= \frac{3}{28}$

$= 10,7$ %
Sorry to bother you again, but I am unsure as to how you have got the:

$\frac{4}{7}$

and

$\frac{2}{7}$...

How were these numbers acquired?

4. Originally Posted by janvdl
I'm not really good with probability, so excuse me if these are not completely correct.

For 1)

We have 8 counters in total, the chances of drawing a green counter would be 5 out of 8, which is 62,5 %

When the 2nd counter is drawn, there are only 7 counters left.

Chances of drawing a red counter would then be 3 out of 7, which is a chance of 42,9%

The chances that the first one is green AND the second one is red is:

$\frac{5}{8} \times \frac{3}{7}$

$= \frac{15}{56}$

$= 26,8$ %

-------------------

For 2)

The chances of drawing two green counters are:

$\frac{5}{8} \times \frac{4}{7}$

$= \frac{5}{14}$

$= 35,7$ %

The chances of drawing two red counters are:

$\frac{3}{8} \times \frac{2}{7}$

$= \frac{3}{28}$

$= 10,7$ %

So the chances of drawing either a green pair, or a red pair is:

$\frac{5}{14} + \frac{3}{28}$

$= \frac{13}{28}$

$= 46,4$ %
your reasoning is correct, i didn't check the computations. so if those check out, you're good

5. Originally Posted by Chris Down
Sorry to bother you again, but I am unsure as to how you have got the:

$\frac{4}{7}$
we had 8 counters to begin with. the chances of choosing a green one is $\frac 58$. having chosen a green counter, there are now 7 counters left, 4 of which are green, since you took out one of the green ones. now the chances of choosing another green counter is $\frac 47$

and

$\frac{2}{7}$...

How were these numbers acquired?
a similar reasoning holds here

6. Originally Posted by Jhevon
we had 8 counters to begin with. the chances of choosing a green one is $\frac 58$. having chosen a green counter, there are now 7 counters left, 4 of which are green, since you took out one of the green ones. now the chances of choosing another green counter is $\frac 47$
Thanks a lot man. I owe you one.