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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: Probability.........different methods
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[Post New]23 Mar 2008 15:43:28 IST
    Subject: Probability.........different methods
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kishan12 (309)

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The probability that a man can hit a target is 3/4.He tries 5 times.The probability that he will hit at least three times is...
Ans :459/512

Friends i know how to get the ans...but plzz shoe me different approches to solve this sum.

All ans Will be saluted and best one will get 5 salutes...
    
[Post New]23 Mar 2008 15:46:20 IST
    Subject: Re:Probability.........different methods
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akhil_o (2709)

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3 times hit:

5C3*(3/4)3(1/4)2

4 times hit:

5C4*(3/4)4(1/4)1

5 times hit:

(3/4)5

addding all these we get required P

P=5C3*(3/4)3(1/4)2+5C4*(3/4)4(1/4)1+(3/4)5
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[Post New]23 Mar 2008 15:51:13 IST
    Subject: Re:Probability.........different methods
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dheer_07 (241)

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theory:  since every time  prob  of  success is  same  n  all  events  r  independent  it  clearly  indicates that we shud go for  binomial  trials thus
 
p[x>=3]
  p3  + p4  + p5
 
so
 
3 times hit:

5C3*(3/4)3(1/4)2

4 times hit:

5C4*(3/4)4(1/4)1

5 times hit:

(3/4)5

addding all these we get required P

P=5C3*(3/4)3(1/4)2+5C4*(3/4)4(1/4)1+(3/4)5
 
 
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[Post New]23 Mar 2008 15:51:28 IST
    Subject: Probability.........different methods
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kishan12 (309)

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Common friends any other approaches..............
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[Post New]24 Mar 2008 10:53:00 IST
    Subject: Probability.........different methods
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spideyunlimited (3871)

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Binomial distribution :

n = 5
p = success, q = failure
p = 3/4, q = 1/4

Probability of hitting 3 times = 5C3 . p3. q2

Probability of hitting 4 times = 5C4 . p4.q1

Probability of hitting 5 times = 5C5. p5 .q0

Total probability = 5C3 .p3q2   +  5C4 .p4.q1   +   5C5. p5.q0

= p3 ( 5C3. q2 +   5C4. pq +   5C5 .p2)

= (3/4)3 [ 10.(1/4)2 + 5.(3/4).(1/4) + (3/4)2 ]

= 27/ 64 [ 10/16 + 15/16 + 9/16]

= 27/64 [ 34/16]

= 549 / 512
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* Agent 'G' [sniper] - SD-6 (Alliance of Twelve)

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