3. A thoroughly shuffled deck of cards is dealt one card at a time until the sec

Question

3. A thoroughly shuffled deck of cards is dealt one card at a time until the second ace is revealed.
(a) What is the probability that the second ace occurs somewhere in the first half of the deck?
Given that the second ace was the 25th card dealt what is the probability of the following:
(b) the next card is the ace of spades?
(c) the next card is the two of clubs?

(Note: this is a question covering some of the material from Conditional Probability & Independence and/or Random Variables (from "A First Course in Probability" by Ross) so it may be tricky. Please explain in details, will leave the best feedback THANKS)

Explanation / Answer

The probability that the second ace occurs womewhere in the first half of the deck is 1 - the probability that it did not occur in the first half. If it did not occur in the first half, then there were only 0 or 1 aces in the first half. P(0 aces) means that 26 of the 48 non-aces were selected and none of the four aces. There are, of course C(52,26) different ways that the first 26 cards could be selected (this is the denominator for the probability calculation). Then, the probability will equal C(48,26)/C(52,26) = 48!/(26!22!) / (52!/26!26!) = 48!/(26!22!) / (52*51*50*49*48!/26*25*24*23*22!26!) = 26*25*24*23/(52*51*50*49)= 24*23/(4*51*49)=46/833 = 0.0552220888355342. The number of ways that 1 ace can be selected = 4 * C(48,25) (the four is because there are 4 aces which you select one; you can also write this as C(4,1) Then, P(1) = 4 * C(48,25)/C(52,26) = 4 * 48!/(25!23!)/(52!/26!26!) = 4 * 48!/(25!23!)/(52*51*50*49*48!/26*25!26*25*24*23!) = 4*26*26*25*24/(52*51*50*49) = 26*24/(51*49) = 26*8/(17*49) = 208/833 = 0.249699879951981 P(0) + P(1) =(46+208)/833 = 254/833 = 0.304921968787515 Then, the probability that the second ace is in the first half of the cards is 1 - 254/833 = 579/833 = 0.695078031212485. b) At this point, 27 cards remain to be selected. Since 2 aces have been selected, the probability is only 2/4 =1/2 that the ace of spades remains among these 27. Then, the probability of selecting the ace of spades is 1/2 * 1/27 = 1/54 = 0.018518519 c) There are 2 aces left to be selected. Thus, 25 non-aces will be selected. Thus, the probability of selecting the 2 of clubs next equals the probability of selecting a non-ace next * the probability of selecting the two of clubs given that a non-ace is selected (1/48). 25/27 * 1/48 = 25/1296 = 0.0192901234567901

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.