What should be an easy conditional probability question

Ten discs numbered 0 to 9 are placed in a box. Two discs are randomly removed from the
box without replacement. Determine the probability that one disc will have the digit 5
written on it and the other disc will have the digit 6 written on it.
A. 1/45
B. 1/50
C. 1/90
D. 1/100
I did (1/10)*(1/9)=1/90
The answer is apparently A. So where did I go wrong?

Hello, flyingfreedom!

Ten discs numbered 0 to 9 are placed in a box.
Two discs are randomly removed from the box without replacement.
Determine the probability that one disc will have the digit 5 and the other disc will have the digit 6.
A. 1/45
B. 1/50
C. 1/90
D. 1/100

I did (1/10)*(1/9)=1/90
The answer is apparently A. So where did I go wrong?

Click to expand...

You imparted an order to the numbers.

Your calculation represents (for example) the probability that the first is 5 and the second is 6.

Since the order can also be "first 6, then 5", the probability is twice as large.


Another approach . . .

\(\displaystyle \text{There are: }\:{10\choose2} \,=\,45\text{ possible pairs of discs that could be drawn.}\)

\(\displaystyle \text{And there is }one\text{ pair which contains a 5 and a 6.}\)

\(\displaystyle \text{Therefore: }\(\text{5 and 6}) \:=\:\frac{1}{45}\)

On your first draw, the probability of getting a desired number (5 or 6) is 2/10. On the second draw the probability is 1/9. This leads to (2/10)*(1/9).