Probability - Target hit

You are absolutely right.

It is the same as the probability of getting at least one shot on target, which is 1-(0.1)^3 =0.999=99.9%

Another way to solve this problem is to consider cases in which the target is hit on the very first shot OR in the second shot OR in the third Shot =0.9+0.1*0.9+0.1*0.1*0.9=0.999=99.9%

It is logical to say that since the chance of hitting the target is 90% on one go, the chance increases as the shooter is given more chance to have a go at the target, but it will never attain 100% probability even if infinite arrows were shot. However the chance of success will tend to move very close towards 100% if infinite arrows are shot.

The Iceman wrote:You are absolutely right.

It is the same as the probability of getting at least one shot on target, which is 1-(0.1)^3 =0.999=99.9%

Thank you.
I was confused because I thought If probability of hitting target in one hit is 90%, how can it be more than 90% in 3 hits.
But after reading your reply I got it. My above doubt stands valid if the question asks the probability of target hit in all three hits. In that case the probability will be well below 90%

9/10 * 9/10 * 9/10 = 729/1000

nisagl750 wrote:

The Iceman wrote:You are absolutely right.

It is the same as the probability of getting at least one shot on target, which is 1-(0.1)^3 =0.999=99.9%

Thank you.
I was confused because I thought If probability of hitting target in one hit is 90%, how can it be more than 90% in 3 hits.
But after reading your reply I got it. My above doubt stands valid if the question asks the probability of target hit in all three hits. In that case the probability will be well below 90%

9/10 * 9/10 * 9/10 = 729/1000

Yes, the odds will go on stacking against the shooter as you have asked too much from him . If you him to do that in 50 such shots his chance reduces from 90% to 0.52%.

Sapana wrote:It is a good question. Thank you for posting it. but, I do not understand what is difference in calculating 0.9*0.9*0.9 for three hits and calculating it by (1-0.001) [0.001=1/10*1/10*1/10]. I know the answers are different but Can any one please explain the difference in logic?

Hi Sapana,

0.9*0.9*0.9 gives the probability that target is hit all three times (i.e all three balls hit the target , Hit Hit Hit)
&
1-0.001 gives the probability that target is hit (i.e target is hit at least once)

I hope you get it. If you have any doubt, please post it.

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Nisagl750

The question says that The probability of hitting the target in one try is 90%.

but I see that here we are using the fact that probability of hitting the target is 90%. which may of may not be the same as probability of hitting the target in one try (single shot).

Can somebody please explain this situation!!!

jkaustubh wrote:The question says that The probability of hitting the target in one try is 90%.

but I see that here we are using the fact that probability of hitting the target is 90%. which may of may not be the same as probability of hitting the target in one try (single shot).

Can somebody please explain this situation!!!

Hi Kaustubh,

Can you please elaborate a bit on your question. This is what I understand.
If the probability of hitting target in one try is 90% then prob of hitting target in every hit is 90%. It goes on increasing with increase in no. of chances given to the player and goes on decreasing with increase in hit ratio required by the player

i.e If the player is given 10 chances, probability will be much higher than 90% (close to 1)
&
If the player is asked to hit 10 out of 10 chances, the probability is very less.
Did I answer you? Or you have any other doubt which I could not understand?

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nisagl750