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A Bag contains 4 red balls and 6 green balls. 4 balls are drawn at random from the bag without replacement

a) Calculate the probability that

i) all the balls are green;

ii) at least one ball of each colour is drawn;

iii) at least two green balls are drawn given that at least one of each colour is drawn.

b) Are the events "at least two green balls are drawn" and "at least one ball of each colour is drawn" independent? Explain.

is a (i) = 6/10 x 5/9 x 4/8 x 3/7 since its without replacement? i'm stuck on this entire question!

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1 Answer

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Your answer for (i) is correct. Think about it combinatorially. We have $\binom{6}{4}$ ways to choose our green balls, and $\binom{10}{4}$ ways to choose $4$ balls. So we divide out:

$$\frac{ \binom{6}{4} } { \binom{10}{4} }$$

For (ii), we want one of each color. So we have one red ball- there are $\binom{4}{1}$ ways to do this. Then there are $\binom{6}{1}$ ways to choose a green ball. Finally, we have $8$ balls left so we choose $2$ balls: $\binom{8}{2}$. We divide out by the number of ways to choose $4$ balls.

(iii) I would look at this in terms of complementation. We have one red and one green ball. What are the odds of getting two red balls (ie., no more green balls)? Subtract this from $1$.

Edit: So we have one of each color has been drawn. So we have three red balls and 5 green balls left. What are the odds of drawing two red balls? We have:

$$\frac{ \binom{3}{2} }{\binom{8}{2} }$$

We choose $2$ red balls and no green balls, then divide out by the number of ways to draw two more balls.

So this is the complement of your solution. So you should easily be able to get the desired solution by subtracting this from $1$.

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