Question 4 A card is selected from a pack of 52 cards.
(a) How many points are there in the sample space?
(b) Calculate the probability that the card is an ace of spades.
(c) Calculate the probability that the card is (i) an ace (ii) black card.
Class X1 - Maths -Probability Page 404
Answers
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Total number of possible outcomes n(S) = 52
(a) hence, there are 52 points in sample space .
(b) there is one ace on spade
∴ number of favourable outcomes n(E) = 1
∴ P(E ) = n(E )/n(S) = 1/52
(c) there are 4 ace in 52 cards
∴ number of favourable outcomes n(E) = 4
∴ P(E) = n(E)/n(S) = 4/52 = 1/13
(d) There are 26 black in a pack of cards .
∴ number of favourable outcomes n(E) = 26
∴ P(E) = n(E)/n(S) = 26/52 = 1/2
(a) hence, there are 52 points in sample space .
(b) there is one ace on spade
∴ number of favourable outcomes n(E) = 1
∴ P(E ) = n(E )/n(S) = 1/52
(c) there are 4 ace in 52 cards
∴ number of favourable outcomes n(E) = 4
∴ P(E) = n(E)/n(S) = 4/52 = 1/13
(d) There are 26 black in a pack of cards .
∴ number of favourable outcomes n(E) = 26
∴ P(E) = n(E)/n(S) = 26/52 = 1/2
Answered by
3
Step-by-step explanation:
(a) There are 52 cards in a pack.
Number of points in the sample space S = n(S) = 52
(b) Let E be the event of drawing an ace of spades.
There is only one ace of spade n(E) = 1 and n(S) = 52
∴ P(E) = n(E) / n(S) = 1 / 52
(c)
(i) Let F be the event of drawing an ace. There are 4 aces in a pack of 52 cards. n(F) = 4, n(S) = 52
∴ P(F) = n(F) / n(s) = 4 / 52 = 1 / 13
(ii) Let G be the event of drawing a black card. There are 26 black cards. n(G) = 26, n(S) = 52
∴ P(G) = n(G)/n(S) = 26 / 52 = 1 / 2
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