A) \[\frac{43}{100}\]
B) \[\frac{87}{200}\]
C) \[\frac{61}{150}\]
D) \[\frac{83}{150}\]
Correct Answer: D
Solution :
Let \[S=A\cup A',\] Where A contains the multiples of 3 and A? is its compliment. Then both \[x\] and \[y\in A\] or one \[\in A\] and other\[\in A'\]. \[\therefore \] Required probability \[=\frac{\left( ^{33}{{C}_{2}} \right)+\left( ^{33}{{C}_{1}} \right)\left( ^{67}{{C}_{1}} \right)}{^{100}{{C}_{2}}}=\frac{83}{150}\]
You need to login to perform this action.
You will be redirected in
3 sec
