Probabilitry

Md Raihan Uddin wrote:Three canons are firing at a target. Their individual probabilities to hit the target are 0.3, 0.4 and 0.5. What is the probability that none of the canons would hit the target?
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
[spoiler]= 0.21
= C[/spoiler]

Cheers,
Brent

Hi Md Raihan Uddin,

In these types of multi-part probability questions, the details matter a great deal, so you should force yourself to take plenty of notes. In that way, you'll know exactly what each "math step" will be. My guess is that your math skills are just fine, so the issue is really about how you organize your work.

Here, we're told the individual probabilities for three canons to hit a target (.3, .4 and .5, respectively). We're asked for the probability that NONE of the canons hit the target.

From a note-taking standpoint, here is one way that you could write down the question:

(Prob that 1st misses)(Prob that 2nd misses)(Prob that 3rd misses) = ?

From here, you can just input the number and multiply….

(.7)(.6)(.5) = .21

Final Answer: C

GMAT assassins aren't born, they're made,
Rich