0.14 More probability  (Page 5/5)

Using the expected value formula, we get

Since there are $\text{51}\mathrm{C6}=\text{18},\text{009},\text{460}$ combinations of six numbers from a total of 51 numbers, the chance of choosing the winning number is 1 out of 18,009,460. So the expected payoff is

This means that every time a person spends $1 to buy a ticket, he or she can expect to lose 89 cents.

Let $U$ be the event that the door has been unlocked and $L$ be the event that the door has not been unlocked. We illustrate with a tree diagram.

Therefore, $\text{the probability of unlocking the door in at most three tries}=1/4+1/4+1/4=3/4$

We illustrate using a tree diagram.

The probability that we will get two black marbles in the first two tries is listed adjacent to the lowest branch, and it $=\frac{3}{10}$

The probability of getting first black, second white, and third black $=\frac{3}{20}$

Similarly, the probability of getting first white, second black, and third black $=\frac{3}{25}$

Therefore, the probability of getting exactly two black marbles in at most three tries $=\frac{3}{10}+=\frac{3}{20}+=\frac{3}{25}=\frac{57}{100}$

Clearly, $\text{the that at least one of the resistors fails}=1-\text{none of the resistors fails}$ .

It is quite easy to find the probability of the event that none of the resistors fails. We don't even need to draw a tree because we can visualize the only branch of the tree that assures this outcome.

The probabilities that $R_1$ , $R_2$ , $R_3$ will not fail are $\text{.}\text{93}$ , $\text{.}\text{90}$ , and $\text{.}\text{92}$ respectively. Therefore, $\text{the probability that none of the resistors fails}=\left(\text{.}\text{93}\right)\left(\text{.}\text{90}\right)\left(\text{.}\text{92}\right)=\text{.}\text{77}$ .

Thus, $\text{the probability that at least one of them will fail}=1-\text{.}\text{77}=\text{.}\text{23}$ .