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 hunnybee July 14th, 2011 12:35 PM

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 !

 mathman July 14th, 2011 01:13 PM

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)

 badatmath July 14th, 2011 01:15 PM

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.

 badatmath July 14th, 2011 01:15 PM

Re: Probability. A deck of card

The other way works too.

 greg1313 July 14th, 2011 02:26 PM

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.

 soroban July 14th, 2011 08:04 PM

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]

 mathman July 15th, 2011 02:58 PM

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]$=\;\;\frac{2499}{2652} \;\;=\;\;\frac{49}{{52}$

 greg1313 July 15th, 2011 03:37 PM

Re: Probability. A deck of card

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

 greg1313 July 16th, 2011 03:54 PM

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?

 mathman July 17th, 2011 01:00 PM

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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