Uh-Oh! That’s all you get for now. We would love to personalise your learning journey. Sign Up to explore more. Sign Up or Login Skip for now Uh-Oh! That’s all you get for now. We would love to personalise your learning journey. Sign Up to explore more. Sign Up or Login Skip for now Two cards can be drawn from 52 cards in 52C2 ways.∴ n(S) = `""^52"C"_2`Also, the pack of 52 cards consists of 26 red and 26 black cards.Let A be the event that both cards are black. ∴ 2 black cards can be drawn in 26C2 ways. ∴ n(A) = 26C2 ∴ P(A) = `("n"("A"))/("n"("S"))=(""^26"C"_2)/(""^52"C"_2)=(26xx25)/(52xx51)=25/102` Let B be the event that both cards are queens. There are 4 queens in a pack of 52 cards ∴ n(B) = `""^4"C"_2` ∴ P(B) = `("n"("B"))/("n"("S")) =(""^4"C"_2)/(""^52"C"_2)= (4xx3)/(52xx51) = 1/221`There are two black queen cards. ∴ n(A ∩ B) = `""^2"C"_2` = 1 ∴ P(A ∩ B) = `("n"("A" ∩ "B"))/("n"("S")) = 1/(""^52"C"_2)=(1xx2xx1)/(52xx51)` = `1/1326`∴ Required probability = P(A ∪ B) P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = `25/102 + 1/221 - 1/1326` = `325/1326 + 6/1236 - 1/1326` = `330/1326` = `55/221`
Q: Find the range of the data 2, 1, 2, 3, 5, 4, 7, 3, 5, 2, 4. Answer & Explanation Answer: D) 6 Explanation: |