Question

A king and his army will attempt to capture a fortress. The left and right flanks break off from the main group to attack the west and east guard towers. Suppose the left flank has a 60% chance of success and the right flank has a 75% chance of success, independently of one another.

If both flanks capture their respective targets, then the king has a 98% chance of successfully taking the fort. If, however, only the left flank captures its tower, the king has an 80% chance of success; if only the right flank succeeds, the king has a 50% chance. If both flanks fail, then the king’s chance of capturing the fort drops to 20%.

It turns out the king captures the fort. What is the probability that one, and only one, flank was successful (either the left, or the right, but not both)?

Answer 1

L - left flank has been captured

R - right flank has been captured

X - successfully taking the fort

there are 4 cases

(x,y) - x represent win/loss on left flank , y for right

Probability of successfully capturing in each case

1) (W,W)   = 0.6 *0.75 * 0.98

2)(W,L) = 0.6 * (1 - 0.75) * 0.80

3) (L,W) =(1 - 0.6) * 0.75 * 0.5 )

4)(L,L) = 0.4*0.25 * 0.20

we have to find the probability

P(((L-R and R-L)| X)

=( P(W,L) + P(L,W))/P(X)

where P(X) is probability of successfully capturing the fort.

= (0.6 * (1 - 0.75) * 0.80 + (1 - 0.6) * 0.75 * 0.5 )/(0.6 * (1 - 0.75) * 0.80 + (1 - 0.6) * 0.75 * 0.5 + 0.6 *0.75 * 0.98 + 0.4*0.25 * 0.20)

= 0.369357

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