Three cannons are firing at a target.

NandishSS wrote:Three cannons are firing at a target. If their individual probabilities to hit the target are 0.3, 0.4, and 0.5 respectively, what is the probability that none of the cannons will hit the target after one round of fire?

A. 0.06
B. 0.12
C. 0.21
D. 0.29
E. 0.94

P(1st MISSES target) = 1 - P(1st HITS target) = 1 - 0.3 = 0.7
P(2nd MISSES target) = 1 - P(2nd HITS target) = 1 - 0.4 = 0.6
P(3rd MISSES target) = 1 - P(3rd HITS target) = 1 - 0.5 = 0.5
----------------------------

P(none hit the target) = P(1st misses target AND 2nd misses target AND 3rd misses target)
= P(1st misses target) x P(2nd misses target) x P(3rd misses target)
= 0.7 x 0.6 x 0.5
= 0.21

Answer: C

Cheers,
Brent

NandishSS wrote:Three cannons are firing at a target. If their individual probabilities to hit the target are 0.3, 0.4, and 0.5 respectively, what is the probability that none of the cannons will hit the target after one round of fire?

A. 0.06
B. 0.12
C. 0.21
D. 0.29
E. 0.94

[spoiler]OA: C[/spoiler]

HI Experts

Why is below ans incorrect?

P [None] = 1 - P [all hit target]
= 1 - (0.3) (0.4) (0.5)
= 0.94

Thanks
Nandish

P[all shots hit target] is not the complement of P[no shots hit target]

If what is the opposite of "none of the shots hit the target"?
Well, it's can be 1 shot hits target or 2 shots hit target or all 3 shots hit target.
So, if we're going to use the complement here, we must consider all 3 cases.

That is:
P(no shots hit target) = 1 - P(1 shot hits target OR 2 shots hit target OR all 3 shots hit target)
Lots of work!!!

Cheers,
Brent

NandishSS wrote:Three cannons are firing at a target. If their individual probabilities to hit the target are 0.3, 0.4, and 0.5 respectively, what is the probability that none of the cannons will hit the target after one round of fire?

A. 0.06
B. 0.12
C. 0.21
D. 0.29
E. 0.94

We need to calculate the probability that none of the 3 cannons will hit the target. P(cannons not hitting) = 0.7 x 0.6 x 0.5 = 0.21.

Answer: C