@snehal
1. Both are valid.
2. see taocp's sol.
The probability of selecting AB to shoot is 1/3, so is BC,AC
If AB were selected, the probability of hitting the target is (1-
probability of both of them missed) = (1-(1-P(A)(1-P(B)),
similar with case BC and AC.
On Jan 11, 11:58 am, snehal jain wro
@snehal i think for first question both statement are correct.when
you
declare const char * cpp then you can not change the value pointed by
this pointer(using this variable),however you can change where it was
pointing.
Thanks and Regards
Priyaranjan
code-forum.blogspot.com
On Jan 11, 9:58 pm, s
@snehal i think for first question both statement are correct.when you
declare const char * cpp then you can not change the value pointed by
this pointer(using this variable),however you can change where it was
pointing.
On Jan 11, 9:58 pm, snehal jain wrote:
> 1. what is valid in cpp
> char *cp;
(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 wrote:
> it is (p(a)p(b)+p(b)p(c)+p(c)p(a))/3
>
> On Jan 12, 1:51 am, SVIX wrote:
>
>
>
> > anuragh
>
> > assume each can shoot the target everytime...
> > P(A) = 1
> > P(B) = 1
> > P(C) = 1
>
it is (p(a)p(b)+p(b)p(c)+p(c)p(a))/3
On Jan 12, 1:51 am, SVIX 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 p
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 th
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 wrote:
> 1. what is valid in cpp
> char *
