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iVinArrow [24]
4 years ago
8

There are five different colored cards in a hat (red, blue, green, yellow, and black). A card is randomly drawn and not replaced

. Then a second card is drawn. What is the probability that the first card will be red and the second card will be blue?
Mathematics
2 answers:
Mamont248 [21]4 years ago
5 0
The probability that the first card will be red<span>= red cards / total cards = 1/5 = 0.2

</span>the probability that the first card <span>will be blue =</span><span><span> blue cards / total cards = 1/4 = 0.25


1/4 because the cards in the second drawn is 4, not 5 cards</span> </span>



KiRa [710]4 years ago
3 0
|\Omega|=5\cdot4=20\\&#10;|A|=1\\\\&#10;P(A)=\dfrac{1}{20}
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step 2

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3 years ago
If nCr : (n+1)Cr : (n+2)Cr = 1:3:7, find the values of n and r.
Troyanec [42]

The solution is n = 5 and r = 4

Step-by-step explanation:

Given,

nCr : (n+1)Cr : (n+2)Cr = 1:3:7

To find the value of n and r.

Formula

nCr = \frac{n!}{r!(n-r)!}  [ n! means = n.(n-1).(n-2)....3.2.1]

Now,

nCr : (n+1)Cr = 1:3                 and (n+1)Cr : (n+2)Cr = 3:7

or, \frac{n!}{r!(n-r)!} : \frac{(n+1)!}{r!(n+1-r)!} = 1:3         or,  \frac{(n+1)!}{r!(n+1-r)!} : \frac{(n+2)!}{r!(n+2-r)!} = \frac{3}{7}

or,  \frac{n!}{r!(n-r)!} × \frac{r!(n+1-r)!}{(n+1)!} = \frac{1}{3}          or, \frac{(n+1)!}{r!(n+1-r)!} ×\frac{r!(n+2-r)!}{(n+2)!} =  \frac{3}{7}  

or, \frac{n+1-r}{n+1} = \frac{1}{3}                               or, \frac{n+2-r}{n+2} =  \frac{3}{7}

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or, 2n-3r = -2                             or, 4n-7r = -8

Now, by solving

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or, 4n-6r-4n+7r = 4

or, r = 4

From (1) we get,

2n = -2+3(4)

or, 2n = 10

or, n = 5

Hence,

n = 5 and r = 4

8 0
4 years ago
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