Two hunters X and Y aim at a target at the same time. The probability that X hits it is 2/9 while the Y misses the target is 6/7. What is the probability that: i. Only one of them hits it? ii. Two of them miss it? iii. Y only hit it? iv. None of them misses it? v. One of them misses it?

19/63;

2/3;

1/9;

2/63;

19/63

Step-by-step explanation:

Given that :

P(X hits) = 2/9

P(X misses) = 1 - 2/9 = 7/9

P(Y misses) = 6/7

P(Y hits) = 1/7

What is the probability that:

i. Only one of them hits it?

[P(X hits) * P(Y misses)] + [P(X misses) * P(Y hits)]

(2/9 * 6/7) + (7/9 * 1/7)

12/63 + 7 /63 = 19/63

ii. Two of them miss it?

P(X misses) * P(Y misses)

(7/9) * (6/7) = 42/63 = 6/9 = 2/3

iii. Y only hit it?

P(X misses) * P(Y hits)

(7/9) * (1/7) = 7/63 = 1/9

iv. None of them misses it?

P(X hits) * P(Y hits)

(2/9) * (1/7) = 2/63

v. One of them misses it?

[P(X hits) * P(Y misses)] + [P(X misses) * P(Y hits)]

[2/9 * 6/7] + [7/9 * 1/7]

12/63 + 7 /63

= 19/63

i believe the answer is 323 ft

step-by-step explanation: