Binomial distribution

Trials that satisfy the following 1 to 3 are called Bernoulli trials

1. In each trial, the only question is whether or not the event occurs.．
2. Each trial is statistically independent．
3. The probability that the event of interest will occur is constant throughout each trial

Let $p$ be the probability that a certain event $X$ will occur in one trial. In the n-time Bernoulli trial sequence, the probability that the event $X$ will occur exactly $i$ times is expressed by

$$P_{r}(X = i) = \binom{n}{i}p^{i}(1-p)^{n-i}$$

At this time, the probability distribution of $X$ is called the binomial distribution and denoted by $X \sim B(n,p)$．

Exercise 8

1. For the binomial distribution $B(8,0.4), \ B(8,0.2)$, find the probability disribution.

2. Suppose there are 25 sets of plastic models, 2 of which are missing some parts. When the customer arbitrarily chooses three sets, find the probability that they are all complete sets.

3. It is said that $40\%$ will die within a certain period of time when a certain reagent is injected into a mouse. When the reagent is injected into 8 mice, answer the following questions.

(a)

Find the probability that all 8 mice will survive for a certain period of time or longe．

(b)

How many out of eight can survive with a probability of $95\%$?

Answer

1. $p_{r} = P(X = r) = \binom{8}{r}(0.4)^{r}(0.6)^{1-r}$. Then

$$\begin{array}{\vert c\vert c\vert c\vert c\vert c\vert c\vert... ....2322 & 0.1239 & 0.0413 & 0.0079 & 0.0007 \\ \hline \end{array}$$

$p_{r} = P(X = r) = \binom{8}{r}(0.2)^{r}(0.8)^{1-r}$. Then

$$\begin{array}{\vert c\vert c\vert c\vert c\vert c\vert c\vert... ...2 & 0.0011 & 0.0001 & 2.56 \times 10^{-6} \\ \hline \end{array}$$

2. The trial to take out the plastic model is Bernoulli trial． Let $X$ be the number of plastic model sets that are missing parts，Then $X \sim B(3,\frac{2}{25})$．Thus，the probability that all three selected sets are complete sets is

$$P(X = 0) = \binom{3}{0}(\frac{2}{25})^{0}(1 - \frac{2}{25})^{3} = 0.7789$$

3. Let $X$ be the number of mice that cannot survive for a certain period of time after injection and die. Then $X \sim B(8,0.4)$ andwe can use the probability distribution obtained in I．

(a) The probability that all eight will survive for a certain period of time is

$$P(X = 0) = \binom{8}{0}(0.4)^{0}(0.6)^{8} = 0.0168$$

(b) Assuming that $r$ is the number of mice that cannot survive with a probability of $95\%$ or more, the distribution function $F(x)$ of $X$ satisfies

$$F(r) = P(X \leq r) = \sum_{k \leq r} r\binom{8}{k}(0.4)^{k}(0.6)^{8-k} \geq 0.95$$

Using the probability distribution of I, we have $P(X = 0) = 0.168, P(X=1) = 0.0896$. Then $r = 1$ and the number of mice that can survive for $95\%$ is 7 or more