1. ## Probability and Stats

I have two questions that I need help on.
1. Bag M contains 5 white balls and 2 red balls and bag N contains 3 white balls and 4 red balls and you randomly select a ball from Bag M and place it in Bag N. If a white ball is selected from Bag N, what is the probability that a red ball was transferred from Bag M to Bag N.

2. There are 10 horses in a race. A particular bet requires a customer to choose the first three horses in the correct finishing order. If all 10 horses have an equal chance of finishing in any position, determine the probability that a single bet wins.

Thanks Bye
~G

2. Originally Posted by the-G
I have two questions that I need help on.
1. Bag M contains 5 white balls and 2 red balls and bag N contains 3 white balls and 4 red balls and you randomly select a ball from Bag M and place it in Bag N. If a white ball is selected from Bag N, what is the probability that a red ball was transferred from Bag M to Bag N.

2. There are 10 horses in a race. A particular bet requires a customer to choose the first three horses in the correct finishing order. If all 10 horses have an equal chance of finishing in any position, determine the probability that a single bet wins.

Thanks Bye
~G
1) P(Red $^M$)
$=\frac {n(Red)}{n(All)}$
$=\frac {2}{7}$

2) P(Win)
$=\frac {n(Win\quad order)}{n(All\quad order)}$
$=\frac {1}{10P3}$
$=\frac {1}{\frac {10!}{7!}}$
$=\frac {1}{720}$

I'm not sure... Just try...

3. 1) P(Red^M)
=\frac {n(Red)}{n(All)}
=\frac {2}{7}
We have an additional piece of information that we can use: a white ball was selected from bag N.

Before we take the ball out, there is Sengnee's 2/7 chance of there being 5 red balls and 3 white balls (case one), and a 1-2/7=5/7 chance of there being 4 red balls and 4 white balls (case 2).
So taking a white ball out of case one has probability $\frac {3}{5+3} * \frac {2}{7} = \frac {3}{28}$ and the probability of taking a white ball out of case 2 is $\frac {5}{7} * \frac {1}{2} = \frac {5}{14}$
So knowing we took a white ball out, the probability that a red ball was transferred is the probability of taking a white ball out of case one divided by the total probability of taking out a white ball ie. $\frac {3/28} {3/28+5/14} = \frac {3}{13}$

So taking a white ball out of case one has probability $\frac {3}{5+3} * \frac {2}{7} = \frac {3}{28}$ and the probability of taking a white ball out of case 2 is $\frac {5}{7} * \frac {1}{2} = \frac {5}{14}$
So knowing we took a white ball out, the probability that a red ball was transferred is the probability of taking a white ball out of case one divided by the total probability of taking out a white ball ie. $\frac {\frac {3}{28}}{\frac{3}{28}}+\frac {5}{14}} = \frac {3}{13}$