Respuesta :
Answer:
[tex]\frac{9}{35}[/tex].
Step-by-step explanation:
We have been given that captain Umaima has probability 3/7 of hitting the pirate ship.
The pirate only has one good eye, so she hits the Captain's ship with probability 2/5.
The probability that pirate misses the captain's ship will be:
[tex]\text{Probability that pirate misses captain's ship}=\text{1-probability that pirate hits captain's ship}[/tex]
[tex]\text{Probability that pirate misses captain's ship}=1-\frac{2}{5}[/tex]
[tex]\text{Probability that pirate misses captain's ship}=\frac{5}{5}-\frac{2}{5}[/tex]
[tex]\text{Probability that pirate misses captain's ship}=\frac{5-2}{5}[/tex]
[tex]\text{Probability that pirate misses captain's ship}=\frac{3}{5}[/tex]
So the probability that pirate misses captain's ship will be 3/5.
Since both events are independent, so by the multiplication rule of probability we will get,
[tex]\text{Probability that captain hits the pirate ship, but pirate misses captain's ship}=\frac{3}{7}\times \frac{3}{5}[/tex]
[tex]\text{Probability that captain hits the pirate ship, but pirate misses captain's ship}=\frac{3*3}{7*5}[/tex]
[tex]\text{Probability that captain hits the pirate ship, but pirate misses captain's ship}=\frac{9}{35}[/tex]
Therefore, the probability that the Captain hits the pirate ship, but the pirate misses will be [tex]\frac{9}{35}[/tex].
The probability that the Captain hits the pirate ship, but the pirate misses is 9/35
What is chain rule in probability?
For two events A and B, by chain rule, we have:
[tex]P(A \cap B) = P(B)P(A|B) = P(A)P(B|A)[/tex]
where P(A|B) is probability of occurrence of A given that B already occurred.
How to find if two events are independent?
Suppose that two events are denoted by A and B.
They are said to be independent event if and only if:
[tex]P(A \cap B) = P(A)P(B)[/tex]
It is because
[tex]P(A|B) = P(A)\\P(B|A) = P(B)[/tex]
and therefore, using chain rule and above facts gives:
[tex]P(A \cap B) = P(A)P(B)[/tex]
So, we're specified that:
- Probability that Captain hist the pirate ship = 3/7
- Probability that the pirate hits th captain's ship = 2/5
Let we take two events as:
- A = event that captain successfully hits the pirate ship
- B = event that pirate misses from hitting the captain's ship.
P(A) = 3/7 as given.
P(B) = 1 - P(event that pirate hits the captain's ship) = 1 - 2/5 = 3/5
Then, the event that Captain hits the pirate ship, but the pirate misses is the intersection of A and B both.
So, P( Captain hits the pirate ship, but the pirate misses) = P(A∩B)
A and B are independent of each other, so we get:
Thus, using the above fact and the chain rule, we get:
[tex]P(A \cap B) = P(A)P(B|A) = P(A)P(B) = \dfrac{3}{7} \times \dfrac{3}{5} = \dfrac{9}{35}[/tex]
Thus, the probability that the Captain hits the pirate ship, but the pirate misses is 9/35
Learn more about probability here:
brainly.com/question/1210781
