#1, 2, 4, A single card is drawn from a dock of cards. What is the probability t

Question

#1, 2, 4,

A single card is drawn from a dock of cards. What is the probability that it will be (a) an ace of hearts, (b) a heart. (c) either a heart or a diamond. (d) either a red king or any queen, and (e) a face card? A single card is drawn from a deck of cards. What is the probability that it will be (a) either a king or a face card, (b) either a king or a spade, and (c) either a heart or a face card? Two cards are drawn in succession from a deck of cards. The first card is replaced and the deck is reshuffled before the second card is drawn. What is the probability that the two cards are (a) both kings, (b) an ace on the first draw and a king on the second draw, and (c) an ace and a king in either order? Consider the same experiments and desired outcomes as in Problem 12-3. but assume that the first card is not replaced before drawing the second card. Repeat the p ability evaluations. Consider the same unreliable switches of Example 12-6, but assume that three are connected in a serial arrangement, that is, all three have to close to make the system Determine the probability of success.

Explanation / Answer

1. (i)P(Ace of hearts ) = 1/52
(ii) P( hearts ) = 13/52 =1/4
(iii) P( heart/ diamond ) = 26/52 =1/2
(iv)P( red king /queen) = 4/52 = 1/13
(v) P(face card ) = 12/52 = 3/13

2.
(i) P(face card /King) = (9+1)/52 = 5/26
(ii)P(spade or king) = ( 13+3)/52 = 4/13
(iii)P(heart/facecard) = (13+9)/52 = 11/26

4.(i) P(king) = 4/52 , if not replaced , propbability of king = 3/51,
so net P = 1/13*1/17 = 1/221
(ii) P(ace) =1/13 , if not replaced , propbability of king = 4/51,
so net P =1/13*4/51 = 4/663
(iii) P(ace then king) = 4/663 , P(king then ace) = 4/663
as either of the case is required ,
hence net P = 2*4/663 = 8/663

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