# Find the number of shots required for the probability to hit the target becomes 0.9, if the chance of hitting the target is 0.4 per shot.?

What I tried to do is this:

I formed a series which describes the probability the gun will hit the target after each shot:

0.4 +0.6*0.4 +0.6*0.6*0.4 etc etc

But I do not know what method I can use to determine the LEAST number of terms which will add up to 0.9 or more.

Relevance

sum of n terms of a geometric series 1+ r + r^2 + r^3 + ....+ r^n = (1-r^n) /(1-r)

H = hits the target

H' = does not hit the target

P(H) + P(H')P(H) + P(H')P(H')P(H) + ..... = 0.9

P(H) = 0.4

.4 + (0.6)(0.4) + (0.6)^2 (0.4) +..... (0.6)^(n-1) (0.4) = 0.9

0.4 ( 1+(0.6)+(0.6)^2 + .... +(0.6)^n ) = 0.9

(0.4) ( 1- (0.6)^n) / (1-0.6) = 0.9

1-(0.6)^n = 0.9

-(0.6)^n = -0.1

(0.6)^n = 0.1

n log(0.6) = log(0.1)

n = log(0.1) /log(0.6) = 4.5 = 5 (rounded)

• Anonymous
6 years ago

chance of missing is .6

chances of having zero hits after go n = .6^n

so after 1 go you have a .6 chance of no hits, then .36 chance of no hits in 2 goes

therefore after n goes, the chance of at least 1 hit is

1 - (chance of no hits)

=1 - .6^n

need 1 - .6^n = .9

.6^n = 1 - .9 = .1

.6^n = .1

n log(.6) = log(.1)

n = log(.1) / log(.6) = 4.51

4.5 shots isn't possible though, but if you remember thaAT we need

1 - .6^n >= .9

so after 4 goes

1 - .6/^4 = .87

after 5 goes

1 - .6/^5 = ..92

1 - .9^4 =

• You do want the probability of MISSING a lot of times to be reduced to 0.1, so you want

(0.6)^N < 0.1. This isn't too hard:

N*log(0.6) < -1

N*(0.778 - 1) < -1

Looks like N needs to be 5.

Check: 0.6^4 = 0.1296 and 0.6^5 = about 0.08. Yes, 5 shots are needed.