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The probability of hitting a Target by 3 marksmen is 1/2,1/3 and 1/4 respectively . Find the probability that one and only one of them will hit the target when they fire simultaneously

Maharshi , 7 Years ago
Grade 12
anser 2 Answers
Vikas TU

Last Activity: 7 Years ago

Dear Student,
As from the given data,
Given individual probabilites are:
P(A) = 3/5 P(A') = 2/5 ( probability of A hitting and not hitting )
P(B) = 2/5 P(B') = 3/5
P(C) = 3/4 P(C') = 1/4
 
Exactly two of them hit:
 
= P( A & B hits & C misses ) + P( A & C hits & B misses ) + P( B & C hits & A misses )
 
= 6/100+27/100+12/100
 
= 45/100
Ans= 1- 45/100
       = 55/100
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)

Sharry

Last Activity: 5 Years ago

P(A)=1/2,P(B)=1/3,P(C)=1/4
=1/2+(1-1/3)+(1-1/4)+(1-1/2)+1/3+(1-1/4)+(1-1/2)+(1-1/3)+1/4
=1/2+2/3+3/4+1/2+1/3+4+1/2+2/3+1/4
=3/2+5/3+7/4 = 11/24
 

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