NIMCET 2007 — Mathematics PYQ

1. Identify the Given Probabilities

Let the probabilities of hitting the target for $A$, $B$, and $C$ be $P(A)$, $P(B)$, and $P(C)$ respectively.

• $P(A)=\frac{3}{5}$

• $P(B)=\frac{2}{5}$

• $P(C)=\frac{3}{4}$

Now, find the probabilities of them missing the target (complements):

• $P(A^{\prime })=1-P(A)=1-\frac{3}{5}=\frac{2}{5}$

• $P(B^{\prime })=1-P(B)=1-\frac{2}{5}=\frac{3}{5}$

• $P(C^{\prime })=1-P(C)=1-\frac{3}{4}=\frac{1}{4}$

2. Understand "At Least Two Shots"

The phrase "at least two shots hit" means the target could be hit by:

1. Exactly two shooters (while the third misses).

2. Exactly three shooters (all of them hit).

Let's break down the possible successful cases:

• Case 1: $A$ and $B$ hit, $C$ misses $⟹P(A\cap B\cap C^{\prime })=P(A)\cdot P(B)\cdot P(C^{\prime })$

• Case 2: $B$ and $C$ hit, $A$ misses $⟹P(A^{\prime }\cap B\cap C)=P(A^{\prime })\cdot P(B)\cdot P(C)$

• Case 3: $A$ and $C$ hit, $B$ misses $⟹P(A\cap B^{\prime }\cap C)=P(A)\cdot P(B^{\prime })\cdot P(C)$

• Case 4: All three ($A$, $B$, and $C$) hit $⟹P(A\cap B\cap C)=P(A)\cdot P(B)\cdot P(C)$

Since these shooters act independently, we can simply multiply the individual probabilities for each case.

3. Calculate the Probabilities for Each Case

• Case 1 ($A,B$ hit; $C$ misses):

  $$P_1=\frac{3}{5}\times \frac{2}{5}\times \frac{1}{4}=\frac{6}{100}$$

• Case 2 ($B,C$ hit; $A$ misses):

  $$P_2=\frac{2}{5}\times \frac{2}{5}\times \frac{3}{4}=\frac{12}{100}$$

• Case 3 ($A,C$ hit; $B$ misses):

  $$P_3=\frac{3}{5}\times \frac{3}{5}\times \frac{3}{4}=\frac{27}{100}$$

• Case 4 (All three hit):

  $$P_4=\frac{3}{5}\times \frac{2}{5}\times \frac{3}{4}=\frac{18}{100}$$

4. Calculate Total Probability

Add the probabilities of all mutually exclusive cases together:

$$\text{Total Probability}=P_1+P_2+P_3+P_4$$

$$\text{Total Probability}=\frac{6}{100}+\frac{12}{100}+\frac{27}{100}+\frac{18}{100}$$

$$\text{Total Probability}=\frac{63}{100}=0.63$$

Correct Option:

(a) $0.63$