0.14 More probability  (Page 2/5)

Let $S$ denote the probability of obtaining a head, and F the probability of obtaining a tail.

Clearly, $n=\text{10}$ , $k=3$ , $p=1/2$ , and $q=1/2$ .

Therefore,

The probability of making a free throw is $3/4$ . Therefore, $p=3/4$ , $q=1/4$ , $n=\text{10}$ , and $k=6$ .

Therefore,

Here $p=\text{.}\text{80}$ , $q=\text{.}\text{20}$ , $n=8$ , and $k=5$ .

If $S$ denotes the probability that the chip is defective, and $F$ the probability that the chip is not defective, then $p=\text{.}\text{04}$ , $q=\text{.}\text{96}$ , $n=\text{60}$ , and $k=3$ .

If S denoted the probability that a person will buy the product, and F the probability that the person will not buy the product, then $p=\text{.}\text{15}$ , $q=\text{.}\text{85}$ , $n=\text{12}$ , and $k=2$ .

$b\left(\text{12},\mathrm{2,}\text{.}\text{15}\right)=\text{12}\mathrm{C2}{\left(\text{.}\text{15}\right)}^2{\left(\text{.}\text{85}\right)}^{\text{10}}=\text{.}\text{2924}$ .

Let $JI$ I be the event that Jar I is chosen, $J\text{II}$ be the event that Jar II is chosen, $B$ be the event that a black marble is chosen and $W$ the event that a white marble is chosen.

We illustrate using a tree diagram.

1. The probability that a black marble is chosen is $P\left(B\right)=1/\text{10}+2/\text{10}=3/\text{10}$ .

2. To find $P\left(JI\mid B\right)$ , we use the definition of conditional probability, and we get

   $P\left(JI\mid B\right)=\frac{P\left(JI\cap B\right)}{P\left(B\right)}=\frac{1/\text{10}}{3/\text{10}}=\frac{1}{3}$

3. Similarly, $P\left(J\text{II}\mid B\right)=\frac{P\left(J\text{II}\cap B\right)}{P\left(B\right)}=\frac{2/\text{10}}{3/\text{10}}=\frac{2}{3}$

   In parts b and c, the reader should note that the denominator is the sum of all probabilities of all branches of the tree that produce a black marble, while the numerator is the branch that is associated with the particular jar in question.