We start with 6 Black playing cards and we add cards to the point where the probability of drawing 2 Red cards in a row without replacement is $\frac{1}{2}$ $1/2$ . How many cards do we add?

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We start with 6 Black playing cards and we add cards to the point where the probability of drawing 2 Red cards in a row without replacement is $\frac{1}{2}$ $1/2$ . How many cards do we add?

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Answer 1 · 2017-02-08T06:59:09

$15$ $15$ red cards.

Explanation:

Let the number of red cards to be added to $6$ $6$ black cards be $x$ $x$ , where $x$ $x$ is natural number.

Now probability of getting first red card is $\frac{x}{x + 6}$ $x/(x+6)$

and conditional probability of getting second red card (assuming we have already got one red card) is $\frac{x - 1}{x + 5}$ $(x-1)/(x+5)$ , as we are considering without replacement.

Hence probability of drawing $2$ $2$ red cards, without replacement, is
$\frac{x}{x + 6} \times \frac{x - 1}{x + 5}$ $x/(x+6)xx(x-1)/(x+5)$ and as this probability is $\frac{1}{2}$ $1/2$ , we have

$\frac{x (x - 1)}{(x + 6) (x + 5)} = \frac{1}{2}$ $(x(x-1))/((x+6)(x+5))=1/2$

or $2 x (x - 1) = (x + 6) (x + 5)$ $2x(x-1)=(x+6)(x+5)$

or $2 x^{2} - 2 x = x^{2} + 11 x + 30$ $2x^2-2x=x^2+11x+30$

or $x^{2} - 13 x - 30 = 0$ $x^2-13x-30=0$

or $(x - 15) (x + 2) = 0$ $(x-15)(x+2)=0$

i.e. $x = 15$ $x=15$ or $x = - 2$ $x=-2$

But as number of red cards can only be a natural number

$x = 15$ $x=15$

Answer 2 · 2017-02-08T07:02:18

If we only add Red cards, 15. If we don't limit the colour of cards added, there is no number that can be added that will achieve that probability.

Explanation:

Let me first point out that if we're talking about solely red cards being added, the answer is 15 and the details are here:

https://socratic.org/questions/how-many-red-cards-must-be-added-to-6-black-cards-so-that-the-probability-of-dra?source=search

That said, this question does not limit itself to red cards being added, and so we can have both black and red added. If we don't limit ourselves to a single pack of cards and can instead draw from an infinite supply of mixed cards , we can approach the question this way - first I'll lay out the starting argument the same way the question handling adding solely red cards does:

We have (B)lack cards and (R)ed cards. We start with $B = 6$ $B=6$ and want to add R so that the odds of drawing 2 R, without replacement, is $\frac{1}{2}$ $1/2$ . How many R must we add?

The odds of drawing an R on a single draw is:

$\frac{R}{R + B}$ $R/(R+B)$

and so for example if we have $R = B = 6$ $R=B=6$ , we'd have the odds of drawing an R to be:

$\frac{6}{6 + 6} = \frac{6}{12} = \frac{1}{2}$ $6/(6+6)=6/12=1/2$

So now let's add that second draw into the mix. We've already drawn an R and so we have one R less, so the ratio for the second draw is:

$\frac{R - 1}{(R - 1) + B}$ $(R-1)/((R-1)+B)$

Which means that the two draws taken together are:

$(\frac{R}{R + B}) (\frac{R - 1}{(R - 1) + B})$ $(R/(R+B))((R-1)/((R-1)+B))$

Now in the analysis with working with only adding R, we know that $B = 6$ $B=6$ . We don't know that here. We know that $B \geq 6$ $B >=6$ and that, in general, the number of R we add will be equal to the number of B added, so we can express that as:

$B = 6 + R$ $B=6+R$

We still want the odds of the two draws to be $\frac{1}{2}$ $1/2$ , so we'll get:

$(\frac{R}{R + 6 + R}) (\frac{R - 1}{(R - 1) + 6 + R}) = \frac{1}{2}$ $(R/(R+6+R))((R-1)/((R-1)+6+R))=1/2$

$(\frac{R}{2 R + 6}) (\frac{R - 1}{2 R + 5}) = \frac{1}{2}$ $(R/(2R+6))((R-1)/(2R+5))=1/2$

$\frac{R (R - 1)}{(2 R + 6) (2 R + 5)} = \frac{1}{2}$ $(R(R-1))/((2R+6)(2R+5))=1/2$

$\frac{R^{2} - R}{4 R^{2} + 22 R + 30} = \frac{1}{2}$ $(R^2-R)/(4R^2+22R+30)=1/2$

$2 (R^{2} - R) = 4 R^{2} + 22 R + 30$ $2(R^2-R)=4R^2+22R+30$

$2 R^{2} - 2 R = 4 R^{2} + 22 R + 30$ $2R^2-2R=4R^2+22R+30$

$0 = 2 R^{2} + 24 R + 30$ $0=2R^2+24R+30$

And now I'll use the Quadratic Formula:

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = (-b \pm sqrt(b^2-4ac)) / (2a)$

$x = \frac{- 24 \pm \sqrt{24^{2} - 4 (2) (30)}}{2 (2)}$ $x = (-24 \pm sqrt(24^2-4(2)(30))) / (2(2))$

$x = \frac{- 24 \pm \sqrt{576 - 240}}{4}$ $x = (-24 \pm sqrt(576-240)) / 4$

$x = \frac{- 24 \pm \sqrt{336}}{4}$ $x = (-24 \pm sqrt(336)) / 4$

$x ≅ \frac{- 24 \pm 18.33}{4}$ $x~= (-24 \pm 18.33) / 4$

And all we end up with is negative results. This means there are no number of mixed cards that will give us the probability of drawing 2 R in a row with the odds of $\frac{1}{2}$ $1/2$ .

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We start with 6 Black playing cards and we add cards to the point where the probability of drawing 2 Red cards in a row without replacement is $\frac{1}{2}$ $1/2$ . How many cards do we add?

2 Answers

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Explanation:

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We start with 6 Black playing cards and we add cards to the point where the probability of drawing 2 Red cards in a row without replacement is $\frac{1}{2}$ $1/2$ . How many cards do we add?