(1-(1-p(a))(1-p(b)) + 1-(1-p(b))(1-p(c)) + 1- (1-p(a))(1-p(c)))/3 On 1月12日, 上午9时36分, ankit agarwal <ankit.agarwal.n...@gmail.com> wrote: > it is (p(a)p(b)+p(b)p(c)+p(c)p(a))/3 > > On Jan 12, 1:51 am, SVIX <saivivekh.swaminat...@gmail.com> wrote: > > > > > 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?- 隐藏被引用文字 - > > - 显示引用的文字 -
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