Latest Infosys Aptitude Question SOLUTION: the probability of hitting a target is 3/4 and a man tried to hit this 5 times.what is the probability of hitting a target atleast thrice?

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the probability of hitting a target is 3/4 and a man tried to hit this 5 times.what is the probability of hitting a target atleast thrice?

Asked In Infosys namrata paul (11 years ago)

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arvind  (9 years ago)

using binomial theoram ncr q^(n-r)p^r
n=5 p=3/4 q=1/4
now atleast three so 5c3 1/4^(5-3)3/4^3 +5c4 1/4^(5-4)3/4^4 +5c5 1/4^(5-5)3/4^5
on solving this equation getting answer
answer is 459/512

Is this solution Helpfull? Yes (19) | No (2)  

priyanshi trivedi  (10 years ago)

hitting target atleast 3 means 3times+4times+5 times.... so (3/4)^3*(1/4)^2+(3/4)^4(1/4)+(3/4)^5(1/4)^0 so ans is .1648

Is this solution Helpfull? Yes (12) | No (8)  

jyotish r  (11 years ago)

THIS IS A BINOMIAL DISTRIBUTION PROBLEM
probability of x successes in n trials is given asP(x)=nCx(p^x)q^(n-x)
where x-no. of success
n- number of attempts
p- probability of success
1-P(0)-P(1)-P(2)=1-.25^5-5*.75*.25^4-10*.75^2*.25^3=.896 IS THE ANSWER

Is this solution Helpfull? Yes (5) | No (10)  

Santanu Panja  (9 years ago)

this is a question on binomial theorem of probability
P(k success out of n trials)= nCk* p^k * q^(n-k)
where
p=probability of success
q=probability of failure

now in this question probability of success p=3/4 then probability of failur q=1-3/4=1/4
therefore P(3 success out of 5 trials)=5C3* 3/4^3 * 1/4^2=135/512
now it asks for at least 3 success then there may be 4 or 5 success also
so, the required probability=P(3 success)+P(4 success)+P(5 success)
P(at least 3 success)=.896

Is this solution Helpfull? Yes (5) | No (1)  

Aman Agarwal  (10 years ago)

The probability of hitting the target is 3/4.
Therefore the probability of missing the target is 1-3/4=1/4.
Hence the probability of hitting the target atleast thrice is 1-(1/4)^2.

Is this solution Helpfull? Yes (4) | No (1)  

R.K.Saipujith Reddy  (9 years ago)

) Let the probability of hitting the target be p = 3/4
So probability of not hitting the target = q = 1 - 3/4 = 1/4
No. of trials = n = 5
Number of successes required = atleast 3; ==> r = 3, 4 & 5

ii) By Binomial distribution probability of r successes in n trials is given by:
P(x = r) = C(n, r)*{p^r}*{q^(n - r)}

Applying the above, P(r = 3) = C(5,3)*{(3/4)^3}*{(1/4)^(2} = 270/(4^5)
Similarly P(r = 4) = C(5,4)*{(3/4)^4}*(1/4) = 405/(4^5)
P(r = 5) = C(5,5)*{(3/4)^5} = 243/4^5

Thus total probability = (270 + 405 + 243)/(4^5) = 459/512

Is this solution Helpfull? Yes (3) | No (1)  

Shafu Gupta  (11 years ago)

ans)2880
4 boys can occupy 4positions hence can be arranged in !4ways(_b1_b2_b3_b4_)
4 girls can be arranged in !5ways because there are 5 positions for girls.
so total no.of ways are= !4*!5= 24*120= 2880.

Is this solution Helpfull? Yes (2) | No (7)  

SHOBANA V  (8 years ago)

i) Let the probability of hitting the target be p = 3/4
So probability of not hitting the target = q = 1 - 3/4 = 1/4
No. of trials = n = 5
Number of successes required = atleast 3; ==> r = 3, 4 & 5

ii) By Binomial distribution probability of r successes in n trials is given by:
P(x = r) = C(n, r)*{p^r}*{q^(n - r)}

Applying the above, P(r = 3) = C(5,3)*{(3/4)^3}*{(1/4)^(2} = 270/(4^5)
Similarly P(r = 4) = C(5,4)*{(3/4)^4}*(1/4) = 405/(4^5)
P(r = 5) = C(5,5)*{(3/4)^5} = 243/4^5

Thus total probability = (270 + 405 + 243)/(4^5) = 459/512

Is this solution Helpfull? Yes (2) | No (1)  

avni sharma  (11 years ago)

P(0 HIT)=1/256
P(1 HIT)=15/256
p(2 HIT)=90/256
P(ATLEAST 3 HIT)=1- 1/256 - 15/256 -90/256= 75/128

Is this solution Helpfull? Yes (1) | No (5)  

rajendra das  (7 years ago)

answer is -459/512

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avinash  (5 years ago)

Probability of hitting a target = Probability of success, p = 34

Probability of not hitting = Probability of failure, q = 1−34=14

Probability of hitting a target at least 3 times out of 5 times = sum of probabilities of hitting a target exactly 3 times, exactly 4 times and exactly 5times

P(X≥3) = P(3) + P(4) + P(5)

= 5C3(34)3(14)2+5C4(34)4(14)1+5C5(34)5

= 10(2764)(116)+5(9256)(14)+1(2431024)

= 2701024+451024+2431024

= 5581024

= 279512

= 0.5449

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