Question

1. Raymond is a basketball player who takes four independent free throws with 70% probability of getting a basket on each shot. Let X be the number of baskets Raymond
gets. Find the probability that he gets exactly 2 baskets, to 3 decimal places.

2. The lifetime of a machine is continuous on the interval (0, 40) with probability density
function f, where f(t) is proportional to (t+ 10)^−2
, and t is the lifetime in years. Find
the proportionality constant that makes the f(t) a probability function.

3. Suppose John throws a die repeatedly until he gets a six. What is the probability that he needs to throw more than 10 times to get a six, to 3 decimal places?

4.A random variable X has Poisson distribution with mean equal to 0.4. What is the probability that the random variable is greater than zero?

Answer 1

$1. \ X\sim Bin(n, p)$

$$P(X=x)=\binom{n}{k}p^x(1-p)^{n-x}$$

Given $p=0.7,n=4$

$$P(X=2)=\binom{4}{2}(0.7)^2(1-0.7)^{4-2}=$$ $$={4!\over 2!(4-2)!}(0.21)^2=0.2646\approx0.265$$

2. Since $f(t)$ is proportional to $(10+t)^{-2}$ on the interval $(0,40),$ we conclude that there is a positive constant $C$ such that

$$f(t) = \begin{cases} C(10+t)^{-2} &\text{if } t\in(0,40) \\ 0, &\text{otherwise} \end{cases}$$

We can determine the proportionality constant $C$ from the condition $\displaystyle\int_{-\infin}^{\infin}f(t)dt=1,$

which must be satisfied because $f(t)$ is a density. We have

$$1=\displaystyle\int_{-\infin}^{\infin}f(t)dt=\displaystyle\int_{0}^{40}C(10+t)^{-2}dt=$$ $$=-C\bigg[{1\over 10+t}\bigg]\begin{matrix} 40 \\ 0 \end{matrix}=C(-{1\over 50}+{1\over 10})={2\over 25}C$$

Therefore $C=12.5.$

$$f(t) = \begin{cases} 12.5(10+t)^{-2} &\text{if } t\in(0,40) \\ 0, &\text{otherwise} \end{cases}$$

3. Geometric distribution

If repeated independent trials can result in a success with probability $p$ and a failure with probability $q=1-p,$ then the probability distribution of the random variable $X,$ the number of the trial on which the first success occurs, is

$$g(x;p)=p(1-p)^{x-1}, x=1,2,3,...$$

Given $p=\dfrac{1}{6}$

$$P(X=1)={1 \over 6}(1-{1 \over 6})^0={1 \over 6}$$ $$P(X=2)={1 \over 6}(1-{1 \over 6})^1={1 \over 6}({5 \over 6})^1$$ $$...$$ $$P(X=10)={1 \over 6}(1-{1 \over 6})^{9}={1 \over 6}({5 \over 6})^{9}$$

$$P(X\leq10)=\displaystyle\sum_{i=1}^{10}P(X=x_i)={1 \over 6}\displaystyle\sum_{i=1}^{10}({5 \over 6})^{i-1}=$$ $$={1 \over 6}\cdot\dfrac{1-({5 \over 6})^{10}}{1-{5\over 6}}=1-({5 \over 6})^{10}$$ $$P(X>10)=1-P(X\leq10)=({5 \over 6})^{10}\approx0.162$$

4.

$$P(X=x)={e^{-\mu}\cdot \mu^x \over x!}$$

Given $\mu=0.4$

$$P(X=0)={e^{-0.4}\cdot 0.4^0 \over 0!}=e^{-0.4}$$

$$P(X>0)=1-P(X=0)=1-e^{-0.4}\approx0.330$$