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EleoNora [17]
3 years ago
12

There are 52 playing cards in a standard deck of cards. If Patrica selects a card at random, what is the probability that it wil

l be a black queen then a red card without replacing the first card.
Mathematics
1 answer:
Alisiya [41]3 years ago
8 0
Since there are two black queens out of 52 cards, there is a 2/52 chance of drawing a black queen first.  This is equivalent to a 1/26 chance.

Now that we have removed a black queen, there are 51 cards left in the deck.  26 of them are red because we only took away a black card.  This means that there is a 26/51 of drawing a red card next.

In order to find the probability of both of these happening, we multiply the two together.  1/26 * 26/51 = 26/1326.  This reduces to 1/51.  So, there is a 1/51 chance of drawing a black queen, then a red card.
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Answer:  Option 'C' is correct.

Step-by-step explanation:

Since we have given that

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Now, we will find the quotient by factoring the numerator:

\mathrm{Use\:the\:rational\:root\:theorem}\\a_0=4,\:\quad a_n=6\\\\\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:2,\:4,\:\quad \\\mathrm{The\:dividers\:of\:}a_n:\quad 1,\:2,\:3,\:6\\\\\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:2,\:4}{1,\:2,\:3,\:6}\\\\-\frac{2}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+2\\\\=\left(x+2\right)\frac{6x^4+15x^3+10x^2+10x+4}{x+2}\\\\=\frac{6x^4+15x^3+10x^2+10x+4}{x+2}=6x^3+3x^2+4x+2\\\\

Now, we will factor it again:

=\left(6x^3+3x^2\right)+\left(4x+2\right)\\\\=2\left(2x+1\right)+3x^2\left(2x+1\right)\\\\=\left(2x+1\right)\left(3x^2+2\right)

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