IPMAT 2025 Exam Solved Papers

According to question, probability of A hitting the target $P(A)=0.4$probability of B hitting the target $P(B)=0.6$So, probability of A missing the target $P\left(A^{c}\right)=0.6$probability of B missing the target $P\left(B^{c}\right)=0.4$Let, E denote the event of A hitting the target first.The event E can occur in the following ways:i.) A hits the target in the first shotii.) A and B both miss the target in first shot and after that A hits in the second shotiii.) A and B both miss the target in first and second shot and after that A hits in the third shot and so on....So basically, $P(E)=P(A)+P\left(A^{c} \cap B^{c} \cap A\right)+P\left(A^{c} \cap B^{c} \cap A^{c} \cap B^{c} \cap A\right)+\ldots \ldots$.or, $P(E)=P(A)+P\left(A^{c}\right) \cdot P\left(B^{c}\right) \cdot P(A)+P\left(A^{c}\right) \cdot P\left(B^{c}\right) \cdot P\left(A^{c}\right) \cdot P\left(B^{c}\right) \cdot P(A)+\ldots \ldots$or, $P(E)=0.4+0.6 \times 0.4 \times 0.4+0.6 \times 0.4 \times 0.6 \times 0.4 \times 0.4+\ldots \ldots$or, $P(E)=0.4(1+0.6 \times 0.4+0.6 \times 0.4 \times 0.6 \times 0.4+\ldots \ldots)$or, $P(E)=0.4\left(1+0.24+0.24^{2}+\ldots \ldots\right)$Now, the term in bracket is an infinite G.P. series with common ratio 0.24or, $P(E)=\frac{0.4}{1-0.24}=\frac{0.4}{0.76}=\frac{40}{76}=\frac{10}{19}$