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 July 14th, 2011, 12:35 PM #1 Member   Joined: Mar 2011 Posts: 42 Thanks: 0 Probability. A deck of card A deck of card is shuffled. What is the chance that the top card is not the jack of clubs or the bottom card is the queen of spades? Thank you !
 July 14th, 2011, 01:13 PM #2 Global Moderator   Joined: May 2007 Posts: 6,784 Thanks: 707 Re: Probability. A deck of card Set up two disjoint events and sum probabilities. Event 1, top card is not jack of clubs - prob = 51/52 Event 2, top card is jack of clubs and bottom card is queen of spades - prob = (1/52)(1/51) Add together.
 July 14th, 2011, 01:15 PM #3 Newbie   Joined: Jul 2011 Posts: 12 Thanks: 0 Re: Probability. A deck of card When you have an or statement like that, there are three possibilities: 1) The top card is not the jack of clubs AND the bottom card is the queen of spades 2) The top card is not the jack of clubs and the bottom card is not the queen of spades 3) The top card is the jack of clubs and the bottom card is the queen of spades. This is a lot of probabilities to keep track of, so one way to do it is to find the probability that the or statement is NOT true and subtract that from 1. The statement is not true when the top card is the jack of clubs and the bottom card is not the queen of spades.
 July 14th, 2011, 01:15 PM #4 Newbie   Joined: Jul 2011 Posts: 12 Thanks: 0 Re: Probability. A deck of card The other way works too.
 July 14th, 2011, 02:26 PM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,949 Thanks: 1141 Math Focus: Elementary mathematics and beyond Re: Probability. A deck of card P(j, q) = (1/52)*(1/52) P(nj, q) = (51/52)*(1/52) P(j, nq) = (1/52)*(51/52) P(nj, nq) = (51/52)*(51/52) P(j, q) + P(nj, q) + P(j, nq) + P(nj, nq) = 1. P(j, q) + P(nj, q) + P(nj, nq) = 2653/2704.
July 14th, 2011, 08:04 PM   #6
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Re: Probability. A deck of card

Hello, hunnybee!

Quote:
 A deck of card is shuffled. What is the probability that the top card is not the $J\clubsuit$ or the bottom card is the $Q\spadesuit?$

Let $T$ = Top,[color=beige] .[/color]$B$ = Bottom.

Formula:
$P\bigg([T,\sim J\clubsuit]\,\vee\,[B,\,Q\spadesuit]\bigg) \;=\;P(T,\sim J\clubsuit)\,+\,P(B,\,Q\spadesuit) \,-\,P\left([T,\sim J\clubsuit]\,\wedge\,[B,\,Q\clubsuit]\right)$

[color=beige]. . . . . . . . . . . . . . . . . . [/color]$=\;\;\;\;\;\;\;\; \frac{51}{52} \;\;\;\;\; + \;\;\;\;\;\; \frac{1}{51} \;\;\;\;\; - \;\;\;\;\;\;\;\;\left(\frac{50}{52}\,\cdot\,\frac{ 1}{51}\right)$

[color=beige]. . . . . . . . . . . . . . . . . . [/color]$=\;\;\frac{2603}{2652}$

edit: corrected silly error
[color=beige] .[/color]

July 15th, 2011, 02:58 PM   #7
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Re: Probability. A deck of card

Quote:
Originally Posted by soroban
Hello, hunnybee!

Quote:
 A deck of card is shuffled. What is the probability that the top card is not the $J\clubsuit$ or the bottom card is the $Q\spadesuit?$

Let $T$ = Top,[color=beige] .[/color]$B$ = Bottom.

Formula:
$P\bigg([T,\sim J\clubsuit]\,\vee\,[B,\,Q\spadesuit]\bigg) \;=\;P(T,\sim J\clubsuit)\,+\,P(B,\,Q\spadesuit) \,-\,P\left([T,\sim J\clubsuit]\,\wedge\,[B,\,Q\clubsuit]\right)$

[color=beige]. . . . . . . . . . . . . . . . . . [/color]$=\;\;\;\;\;\;\;\; \frac{51}{52} \;\;\;\;\; + \;\;\;\;\;\; \frac{1}{51} \;\;\;\;\; - \;\;\;\;\;\;\;\;\left(\frac{50}{52}\,\cdot\,\frac{ 1}{51}\right)$

[color=beige]. . . . . . . . . . . . . . . . . . [/color]$=\;\;\frac{2499}{2652} \;\;=\;\;\frac{49}{{52}$

July 15th, 2011, 03:37 PM   #8
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Re: Probability. A deck of card

Quote:
 Originally Posted by mathman . . . - prob = (1/52)(1/51)
mathman, did you intend (1/52)(1/52) ?

July 16th, 2011, 03:54 PM   #9
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Re: Probability. A deck of card

Quote:
 Originally Posted by greg1313 P(j, q) = (1/52)*(1/52) P(nj, q) = (51/52)*(1/52) P(j, nq) = (1/52)*(51/52) P(nj, nq) = (51/52)*(51/52) P(j, q) + P(nj, q) + P(j, nq) + P(nj, nq) = 1. P(j, q) + P(nj, q) + P(nj, nq) = 2653/2704.

I guess this is wrong, but can anyone tell me why?

July 17th, 2011, 01:00 PM   #10
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Re: Probability. A deck of card

Quote:
Originally Posted by greg1313
Quote:
 Originally Posted by mathman . . . - prob = (1/52)(1/51)
mathman, did you intend (1/52)(1/52) ?
No. The prob that the jack of clubs is on top is 1/52. The probability the the queen of spades is on the bottom (given the jack of clubs on top) is 1/51.
Note to soroban: The last term should be (51/52)(1/51)

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