A card is drawn at random from a well shuffled deck of 2 cards. Find the probability of its being a spade or a king.

Correct Answer - C `(c )` Hence the sum of the numbers on six cards vanishes. Case I : If selected `3` cards each of number `-1` or `1` i.e. Number...

Correct Answer - C `(c )` `P(A)=(3)/(4)`, `P(B//A)=(1)/(4)` `P(A//B)=(2)/(3)` `P(B//A)=(P(BnnA))/(P(A))=(1)/(4)` (given) `:.P(BnnA)=(3)/(4)*(1)/(4)=(3)/(16)` Now `P(A//B)=(P(AnnB))/(P(B))=(2)/(3)`(given) `:.P(B)=(3)/(2)*(3)/(16)=(9)/(32)` `:. P(AuuB)=(3)/(4)+(9)/(32)-(3)/(16)=(24+9-6)/(32)=(27)/(32)` `:. P(A^(C )nnB^(C ))=1-P(AuuB)=1-(27)/(32)=(5)/(32)`

Correct option is  (d) Takshaka

Correct Answer - `13/204`

Correct Answer - `"Mean "=(6)/(4), "variance "=(30)/(49)`

Correct Answer - ` (i) 26/49 (ii) 1/49 (iii) 2/49 (iv) 1/49 `

Correct Answer - ` (i) 1/10 (ii) (a) 9/10 (b) 1/5`

Correct Answer - `(i) 9/20 (ii) 7/20 (iii) 4/5 (iv) 11/20 (v) 13/20`

Let X be the random variable defined as the number of red balls. Then X = 0, 1 P(X = 0) = \(\frac{3}{4}\times\frac{2}{3}=\frac{6}{12}=\frac{1}{2}\) P(X = 1) = \(\frac{1}{4}\times\frac{3}{3}+\frac{3}{4}\times\frac{1}{3}=\frac{6}{12}=\frac{1}{2}\) Probability Distribution Table: X 0 1 P(X) \(\frac{1}{2}\) \(\frac{1}{2}\)

The required probability = P((The first is a red jack card and The second is a jack card) or (The first is a red non-jack card and The second is...