# Thread: Probabilities - more questions

1. ## Probabilities - more questions

3. A bag contains 10 balls, 4 are red and 6 are yellow. Three balls are drawn without replacement. What is the probability of drawing at least two red balls?

4. A town has two fire engines operating independently. The probability that a specific fire engine is available when need is 0.95. What is the probability that neither is available when needed? What is the probability that a fire engine is available when needed?

OK could someone help out a little bit my brain is failing to work, I have done some working out but none of it seems to make sense and it seems easy. Don't know what's going on up there.

2. What is the working out you have done?

3. Hello, djmccabie!

3. A bag contains 10 balls, 4 are red and 6 are yellow.
Three balls are drawn without replacement.
What is the probability of drawing at least two red balls?

We draw 3 balls from 10 balls.
There are: . ${10\choose3} = 120$ possible outcomes.

We want 3 Red balls, or 2 Red and 1 Yellow.

. . 3R: . ${4\choose3} = 4$ ways.
. . 2R, 1Y: . ${4\choose2}{6\choose1} = 6\cdot6 = 36$ ways.

Hence, there are: . $4 + 36 \:=\:40$ ways to get at least 2 red balls.

Therefore: . $P(\text{at least 2 R}) \;=\;\frac{40}{120} \;=\;\frac{1}{3}$

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You could have solved it taking baby steps . . .

There are four ways to get at least 2 red balls: .RRR, RRY, RYR, YRR.

. . $\begin{array}{ccccc}P(RRR) &=& \frac{4}{10}\cdot\frac{3}{9}\cdot\frac{2}{8} &=& \frac{1}{30} \\ \\[-3mm]
P(RRY) &=&\frac{4}{10}\cdot\frac{3}{9}\cdot\frac{6}{8} &=& \frac{1}{10} \\ \\[-3mm]
P(RY\!R) &=& \frac{4}{10}\cdot\frac{6}{9}\cdot\frac{3}{8} &=& \frac{1}{10} \\ \\[-3mm]
P(Y\!RR) &=& \frac{6}{10}\cdot\frac{4}{9}\cdot\frac{3}{8} &=& \frac{1}{10}\end{array}$

Therefore: . $P(\text{at least 2 R}) \;=\;\frac{1}{30} + \frac{1}{10} + \frac{1}{10} + \frac{1}{10} \;=\;\frac{10}{30} \;=\;\frac{1}{3}$