mallireddy wrote:The probability that A will successfully hit a target is 2/3. The probability of B succesfully hitting a target is 2/3. if they shoot two shots each then what is the probability that they have an equal number of hits?
AND means MULTIPLY.
OR means ADD.
Case 1: Each hits the target 0 times
P(A misses the 1st shot AND the 2nd shot) = 1/3 * 1/3 =
1/9.
P(B misses the 1st shot AND the 2nd shot) = 1/3 * 1/3 =
1/9.
Since a favorable result requires that A miss both shots (1/9) AND that B miss both shots (1/9), we MULTIPLY the fractions in red:
1/9 * 1/9 =
1/81.
Case 2: Each hits the target once
P(A hits the 1st shot AND misses the 2nd shot) = 2/3 * 1/3 =
2/9.
P(A misses the 1st shot AND hits the 2nd shot) = 1/3 * 2/3 =
2/9.
Since a favorable result requires that A hit only the 1st shot (2/9) OR that A hit only the 2nd shot (2/9), we ADD the fractions in red:
2/9 + 2/9 =
4/9.
P(B hits the 1st shot AND misses the 2nd shot) = 2/3 * 1/3 =
2/9.
P(B misses the 1st shot AND hits the 2nd shot) = 1/3 * 2/3 =
2/9.
Since a favorable result requires that B hit only the 1st shot (2/9) OR that B hit only the second shot (2/9), we ADD the fractions in red:
2/9 + 2/9 =
4/9.
Since a favorable result requires that A hit the target exactly once (4/9) AND that B hit the target exactly once (4/9), we MULTIPLY the fractions in blue:
4/9 * 4/9 =
16/81.
Case 3: Each hits the target twice
P(A hits the first shot AND the 2nd shot) = 2/3 * 2/3 =
4/9.
P(B hits the first shot AND the 2nd shot) = 2/3 * 2/3 =
4/9.
Since a favorable result requires that A hit both shots (4/9) AND that B hit both shots (4/9), we MULTIPLY the fractions in red:
4/9 * 4/9 =
16/81.
Since a favorable outcome will be yielded by Case 1 OR Case 2 OR Case 3, we ADD the fractions in green:
1/81 + 16/81 + 16/81 = 33/81 = [spoiler]11/27[/spoiler].
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