A and B take part in a duel. A can strike with an accuracy of 0.6. B c

Solution

• Let us assume that a denotes probability of A to strike the target and \(a ̅ \) denotes the probability that A doesn’t strike the target.

o \(a = 0.6\)
o \(a ̅ = (1-0.6) = 0.4\)
• Similarly, let us assume that b denotes probability of B to strike the target and \(b ̅ \) denotes the probability that B doesn’t strike the target.

o \(b = 0.8\)
o \(b ̅ = (1-0.8) = 0.2\)
• Now, possible ways in which A can win = A strike the target on his first shot OR A strike the target on his second shot OR A strike the target on his third shot OR .....and so on.
• Since, A and B are striking alternately.

o So, the probability of A to win(Let’s say P) \(= a + a ̅*b ̅ *a + a ̅*b ̅ *a ̅*b ̅ *a +a ̅*b ̅ *a ̅*b ̅ *a ̅*b ̅ *a + ……. \)And so on
o \(P = 0.6 + 0.4*0.2*0.6 + 0.4*0.2*0.4*0.2*0.6 + 0.4*0.2*0.4*0.2*0.4*0.2*0.6 + ………… \)and so on.
o \(P = 0.6 [1 + (0.08) + (0.08) ^2 + (0.08) ^3 + ……….]\)
• We can observe that the terms in the bracket [] forms a G.P. with first term as 1 and common ratio as 0.08.
• Therefore, \(P = 0.6* \frac{1}{1- 0.08} = \frac{0.6}{0.92} = \frac{60}{92} = \frac{15}{23}\)