anuragh.... assume each can shoot the target everytime... P(A) = 1 P(B) = 1 P(C) = 1
per your logic, the probability that the target will be hit is 3.... actually, it should have only been 2 as we're going to pick only 2 people out of 3 to shoot... I think you should factor in the probability that A or B or C will be picked... There are 3C2 ways to pick 2 cards out of 3... Since its purely random, each card has 2/3rd chance that it's picked... so if you factor in the probability, the answer is required probablilty = P(A) * 2/3 + P(B) * 2/3 + P(C) * 2/3 On Jan 11, 12:06 pm, "anurag.singh" <anurag.x.si...@gmail.com> wrote: > For 2nd question (probability): Looks like one data is missing for C. > If I assume C can shoot 8 out of 10. times then: > > P(A) = 4/5 > P(B)=6/7 > P(C)=8/10 > > Required Probability should be = P(A) * P(B) + P(B) * P(C) + P(A) * > P(C) > > On Jan 11, 9:58 pm, snehal jain <learner....@gmail.com> wrote: > > > 1. what is valid in cpp > > char *cp; > > const char* cpp; > > 1. cpp=cp 2. cp=cpp > > > 2 there r 3 ppl A B C > > A can shoot the target 4 out of 5 times B can shoot 6 out of 7 times > > and C can shoot 8 out of times. 2 people r selected at random. then > > wat is the probability of hitting the target? -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
