Thank you for visiting 1 Azri and Fahmi are shooting at a target The probability that Azri s shoot hits the target is 1 2 and the probability that. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
To solve this problem, we need to determine probabilities based on given events involving Azri and Fahmi shooting at a target:
Let's define the events:
Let A be the event that Azri hits the target.
The probability that Azri hits the target, [tex]P(A)[/tex], is [tex]\frac{1}{2}[/tex].
Let B be the event that Fahmi hits the target.
We know the probability that Fahmi misses the target is [tex]\frac{1}{3}[/tex], so the probability that Fahmi hits the target, [tex]P(B)[/tex], is [tex]1 - \frac{1}{3} = \frac{2}{3}[/tex].
Now let's solve the questions:
(a) Both their shots hit the target:
The probability of both Azri and Fahmi hitting the target is the product of their individual probabilities of hitting the target, since each shot is independent of the other:
[tex]P(A \cap B) = P(A) \times P(B) = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3}[/tex]
(b) Only one of their shots hits the target:
This can happen in two scenarios:
- Azri hits the target and Fahmi misses.
- Fahmi hits the target and Azri misses.
Calculating each scenario:
Azri hits and Fahmi misses:
[tex]P(A \cap B^c) = P(A) \times (1 - P(B)) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}[/tex]Fahmi hits and Azri misses:
[tex]P(A^c \cap B) = (1 - P(A)) \times P(B) = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3}[/tex]
Thus, the probability that only one of their shots hits the target is:
[tex]P((A \cap B^c) \cup (A^c \cap B)) = P(A \cap B^c) + P(A^c \cap B) = \frac{1}{6} + \frac{1}{3} = \frac{1}{2}[/tex]
(c) None of their shots hit the target:
This means both Azri and Fahmi miss:
[tex]P(A^c \cap B^c) = (1 - P(A)) \times (1 - P(B)) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}[/tex]
Therefore, the probabilities for these events are:
(a) Both hit: [tex]\frac{1}{3}[/tex]
(b) Only one hits: [tex]\frac{1}{2}[/tex]
(c) None hit: [tex]\frac{1}{6}[/tex]
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