
When the missile is fired from the ship, the probability that it is intercepted is $\dfrac{1}{3}$ and the probability that the missile hits the target, given that it is not intercepted, is $\dfrac{3}{4}$. If three missiles are fired independently from the ship, then the probability that all three hit the target, is?
1. $\dfrac{1}{8}$
2. $\dfrac{1}{{27}}$
3. $\dfrac{3}{4}$
4. $\dfrac{3}{8}$
Answer
162.9k+ views
Hint: In this question, we are given the probability of the missile being intercepted and the missile hitting the target, and we have to find the probability of all three independently fired missiles hitting the target. So, calculate the cube of the product of the probability of the missile not being intercepted and the missile hitting the target.
Formula Used:
Probability of event not occurring –
$P\left( {A'} \right) = 1 - P\left( A \right)$, where $P\left( A \right)$ is the probably of occurring an event
Complete step by step Solution:
Given that,
The probability that the missile is intercepted $P\left( I \right) = \dfrac{1}{3}$
As we know, the sum of occurring and not occurring an event is equal to one.
Therefore, the probability that the missile is not intercepted will be $P\left( {I'} \right) = 1 - P\left( I \right)$
$P\left( {I'} \right) = \dfrac{2}{3}$
Also, the probability that the missile hits the target is $P\left( T \right) = \dfrac{3}{4}$
Now, it is given that a missile hits the target only when it is not intercepted
Therefore, the Probability of hitting the target will be the product of the probability of missile not intercepted and the missile hits the target i.e., $P = \dfrac{2}{3} \times \dfrac{3}{4} = \dfrac{1}{2}$
It implies that the probability that all three-missile hit the target is ${\left( P \right)^3} =\mathrm{{(\dfrac{1}{2})}^{3}}= \dfrac{1}{8}$
Hence, the correct option is 1.
Note: To solve such problems, one should have a good understanding skill. As these questions are all based on their wording, not on formulas. Also, one should always remember that the sum of the probability of an event occurring and not occurring is always equal to one.
Formula Used:
Probability of event not occurring –
$P\left( {A'} \right) = 1 - P\left( A \right)$, where $P\left( A \right)$ is the probably of occurring an event
Complete step by step Solution:
Given that,
The probability that the missile is intercepted $P\left( I \right) = \dfrac{1}{3}$
As we know, the sum of occurring and not occurring an event is equal to one.
Therefore, the probability that the missile is not intercepted will be $P\left( {I'} \right) = 1 - P\left( I \right)$
$P\left( {I'} \right) = \dfrac{2}{3}$
Also, the probability that the missile hits the target is $P\left( T \right) = \dfrac{3}{4}$
Now, it is given that a missile hits the target only when it is not intercepted
Therefore, the Probability of hitting the target will be the product of the probability of missile not intercepted and the missile hits the target i.e., $P = \dfrac{2}{3} \times \dfrac{3}{4} = \dfrac{1}{2}$
It implies that the probability that all three-missile hit the target is ${\left( P \right)^3} =\mathrm{{(\dfrac{1}{2})}^{3}}= \dfrac{1}{8}$
Hence, the correct option is 1.
Note: To solve such problems, one should have a good understanding skill. As these questions are all based on their wording, not on formulas. Also, one should always remember that the sum of the probability of an event occurring and not occurring is always equal to one.
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