Question

$A,B,C$ try to hit a target simultaneously but inside- target are $\dfrac{3}{4},\dfrac{1}{2},\dfrac{5}{8},$ the probability that target is bur by $A$ or $B$ but not by $C$ is
(A) $\dfrac{{21}}{{64}}$
(B) $\dfrac{7}{8}$
(C) $\dfrac{7}{{32}}$
(D) $\dfrac{9}{{64}}$

Answer 1

We have given the individual probabilities. The probability of failure or not occurrence is equal to “1 – probability of success or occurrence”. In probability, “or” represents “addition” and “and” represents “multiplication”. Use these concepts to find the possible outcomes of probability and hence, the total outcome.

Complete Step by Step Solution:
Probability of Hit by $(A) = \dfrac{3}{4}.$
Probability of Hit by $(B) = \dfrac{1}{2}.$
Probability of Hit by $(C) = \dfrac{5}{8}.$
We have to find the probability of hit by $A$ and hit by $B$ but not hit by $C$
Let the probability hit by $'A' = P{(A)_H}$
Let the probability not hit by $(A) = P{(A)_{NH}}$
Let the probability hit by $(B) = P{(B)_H}.$
Let the probability not hit by $(B) = P{(B)_{NH}}.$
Let the probability hit by $(C) = P{(C)_{NH}}$
Then ,$P{(A)_H} = \dfrac{3}{4}$
$P{(B)_H} = \dfrac{1}{2}$
$P(C) = \dfrac{5}{8}$
First we have to find the probability not of each.
Then,$P{(A)_{NH}} = \dfrac{1}{4}.$
$P{(B)_{NH}} = \dfrac{1}{2}$
$P{(C)_{NH}} = \dfrac{3}{8}$
Now we have to find the probability of hit by $'A'$ and hit by $'B'$ but not by $'C'$
Let $P(x)$ represents the probability of hit by $'A'$ and hit by $'B'$but not hit by $'C'$
So,$P(x) = P{(A)_H} \times P{(B)_{NH}} \times P{(C)_{NH}} + P{(B)_H} \times P{(A)_{NH}} \times P{(C)_{NH}} + P{(C)_{NH}} \times P{(A)_H} \times P{(B)_H}.$ . . . (1)
Using this formula to find $P(x) = ?$
Therefore $P(x) = \dfrac{3}{4} \times \dfrac{1}{2} \times \dfrac{3}{8} + \dfrac{1}{4} \times \dfrac{1}{2} \times \dfrac{3}{8} + \dfrac{3}{4} \times \dfrac{1}{2} \times \dfrac{3}{8}$
$= \dfrac{9}{{64}} + \dfrac{3}{{64}} + \dfrac{9}{{64}}$
Probability of Hit $'A'$ and hit $'B'$ but not $'C' = \dfrac{{21}}{{64}}$

**Therefore, from the above explanation, the correct answer is (A) $\dfrac{{21}}{{64}}$

Note: **
You need to think of the various possibilities of outcomes while calculating probability. Just like in this question (1) explains the different possibilities of outcomes. And you need to know that “or” represents “addition” and “and” represents “multiplication”. Then only, you will be able to understand which probabilities to add and which to multiply.