combinationsprobability
There are $6$ red balls and $5$ blue balls in a jar. You pick any $4$ balls without looking in the jar. What is the probability that you would be having $4$ red balls in hand?
Note that you're picking up all the $4$ balls in one single attempt and not one-by-one. Also balls of same colour are to be considered as identical.
I tried this and got the answer as $4/11$ which was wrong. I can't figure out what the cases (sample space) would be.
The total number of ways to choose $3$ out of $8$ balls is $\binom83=56$
The number of ways to choose $2$ out of $3$ blue balls and $1$ out of $5$ red balls is $\binom32\cdot\binom51=15$
Hence the probability is $\dfrac{15}{56}$
Three approaches:
(1) This can be viewed as a hypergeometric distribution. The urn contiains four red balls and five blue balls. Let $X$ be the number of red balls among five balls drawn at random without replacement. To draw more blue balls then red you need to evaluate $P(X \le 2).$ In R statistical software this can be evaluated as follows:
phyper(2, 4, 5, 5)
## 0.6428571
(2) The equivalent answer can be obtained using a combinatorial argument:
$$\frac{{4 \choose 0}{5 \choose 5}+{4\choose 1}{5 \choose 4}+{4 \choose 2}{5 \choose 3}}{{9 \choose 5}} = \frac{1 + 20 + 60}{126} = 81/126 = 9/14 = 0.6428571$$
(3) An approximate value (to about 3 places) from simulating a million draws of five balls from such an urn and counting the red balls can be obtained as follows:
set.seed(616)
m = 10^6; urn = c(1,1,1,1,1,2,2,2,2) # 1 = blue, 2 = red
r = replicate(m, sum(sample(urn, 5)==2)) # sample 5 balls without replacement
mean(r <= 2) # mean of logical vector is nr of TRUEs
## 0.642822
The histogram below shows the simulated hypergeometric distribution of the number of red balls drawn. The open red dots show exact hypergeometric probabilities. At the scale of the graph, it is not easy to see any difference between the simulated and exact values.
Note: You should be sure you understand and can explain the details of either method (1) or method (2) for your class. The simulation is probably not something you are expected to know.
Best Answer
You are picking 4 balls without replacement. The probability that you pick four red balls is $$P =\frac{6}{11} \cdot \frac{5}{10}\cdot \frac{4}{9}\cdot \frac{3}{8}\\P=\frac{360}{7920}\\P=\frac{1}{22}\\P \approx 0.045 \\P \approx 4.5\%$$