Question

Julia enjoys shooting paper balls into the wastebasket across her office. She misses the first shot 50% of the time. When she misses on the first shot, she misses the second shot 20% of the time.
Part A: What is the probability of missing two shots in a row?

Part B: What is the probability of making at least one successful throw?

Answer 1

Answer:

a) Probability that Julia misses two shots in a row = $\frac{1}{2}\frac{1}{5} =\frac{1}{10}$

b) We can use the complement rule on this case. Let X the number of shoots successful, we want this probability:

$P(X \geq 1)$

And using the complement rule we have this:

$P(X \geq 1) = 1-P(X$

And P(X=0) correspond to the probability that Julia misses two shots in a row founded on part A so then we have:

$P(X\geq 1) = 1- \frac{1}{10}= \frac{9}{10}$

Step-by-step explanation:

Part a

For this case we know that Julia enjoys shooting paper balls into the basket across her office.

She misses the first shot 50% of the time.

Probability that Julia misses the first shot = $\frac{1}{2}$

When she misses on the first shot, she misses the second shot 20% of the time.

Probability that Julia misses the second shot = $\frac{1}{5}$

For this case we need to find the probability of missing two shots in a row

So we can multiple the two probabilities like this

Probability that Julia misses two shots in a row = $\frac{1}{2}\frac{1}{5} =\frac{1}{10}$

Part b

For this case we want probability of making at least one successful throw

And we can use the complement rule on this case. Let X the number of shoots successful, we want this probability:

$P(X \geq 1)$

And using the complement rule we have this:

$P(X \geq 1) = 1-P(X$

And P(X=0) correspond to the probability that Julia misses two shots in a row founded on part A so then we have:

$P(X\geq 1) = 1- \frac{1}{10}= \frac{9}{10}$