The question is like this, "A bag contains 7 green balls, 4 red balls, and 5 blue balls."
- 5 balls are drawn without being placed back. What is the probability that 2 or more blue balls are drawn?
- Balls are drawn from the bag until a blue ball is found. What is the probability that this will take 3 or more draws?
- 5 balls are drawn and placed back each time. What is the probability function for the number of green balls drawn?
- Same question as 3. but the balls not placed back.
My answers:
- This one I used hyper geometric distribution. So, $F(x \geq 2)=1- (F(x = 0) + F(x = 1))=1 – (\frac{\binom{5}{1} \binom{11}{4}}{\binom{16}{5}}+ \frac{\binom{5}{2} \binom{11}{3}}{\binom{16}{5}})$
- This is finding the first time of choosing a blue ball. So geometric distribution is used here. $F(x \geq 3)=1-(F(x = 0)+F(x = 1)+F(x = 2))$. The equation is too long so I'm not write it. But it's simple, so $F(x = 1)=(0.3125)(0.6875)$ because probability is $\frac{5}{16}=0.3125$. 5 blue balls
- This is using a binomial distribution. So $F(x)=\binom{5}{x}(0.4375)^x(0.5625)^{5-x}$. The probability is $\frac{7}{16}=0.4375$
- This is similar like 3. But I used hyper geometric distribution. $F(x)=\frac{\binom{7}{x}\binom{9}{5-x}}{\binom{16}{5}}$, because there's 16 balls total, and 7 green balls.
Anyone can check my answer is helpful, thank you.
Best Answer
a) Your idea for the first one is right but you have mistakes in what you have written.
$\quad\begin{align}P(X \geq 2) &= 1 - P(X=0) - P(X = 1)\\ &= 1 - \left[\left.{5\choose 0}{11 \choose 5} \middle/ {16 \choose 5}\right. + \left.{5 \choose 1} {11 \choose 4} \middle/ {16 \choose 5}\right. \right]\\& = \frac{47}{91}\end{align}$
b) For the second question, we are to find the probability that it takes more than two draws to get a blue ball. In other words, we do not draw a blue ball in the first two draws. So if we are doing this without replacement, the probability is given by
$ \displaystyle \frac{11}{16} \cdot \frac{10}{15} = \frac{11}{24}$
Third and fourth are correct.