# An urn contains 5 green and 2 red balls.

An urn contains 5 green and 2 red balls. One ball is drawn at random and its colour is recorded. This selected ball is then replaced in the urn and 3 more balls of the same colour are added to the urn. Next, another ball is drawn from the urn and its colour is recorded.

A) Find a suitable sample space for that random experiment and assign probabilities to sample points.

B) Find the probability distribution table of the random variable X representing the number of red balls among the two selected ones. Draw the bar chart of X.

C) Draw the cumulative distribution function F(x).

• Please tell us what your thoughts are, what you've tried, where you're stuck. – rogerl Nov 17 '15 at 23:39

## 1 Answer

The sample space part is straightforward-- they are only drawing two balls, and the only colors are green and red. So the sample space is ${GG, GR, RG, RR}$, with $GR$ denoting green on the first pick, red on the second, etc.

To assign probabilities, \begin{align} P(GG) &= (5/7)*P(G \mbox{ on second pick } | G \mbox{ on first pick})\\ & = (5/7)* (8/10) \end{align} the others are computed similarly.

For (b) just use $P( \mbox{2 green balls}) = P(GG)$, $P(\mbox{1 green ball, 1 red ball}) = P(GR) + P(RG)$, and $P( \mbox{2 red balls}) = P(RR)$ and compute using part (a).

(C) is just drawing a graph of the probabilities from (b).

Voila!

• Can you elaborate more on b?@travelingbones – lamar95 Nov 18 '15 at 1:16
• to calculate P(GR): what's the probability of drawing a green first? (5/7). What's the probability of drawing a red after drawing a green first? (2 reds out of 7+3 more = 10 greens, so 2/10. So the P(GR) = (5/7)*(2/10). You can do the rest. Draw a picture for the first draw, then of the second after adding the appropriate balls and just count – travelingbones Nov 18 '15 at 1:44