danielkim118
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In a sample of 10 cards, 4 are red and 6 are blue. If 2 cards are selected at random from the sample, one at a time without replacement, what is the probability that both cards are not blue?

Respuesta :

The exact answer in fraction from is 2/15

The approximate answer in decimal form is 0.1333 (rounded to four decimal places)

The decimal value 0.1333 is equivalent to 13.33%

These are three ways to say the same basic answer.

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Explanation:

We have 4 red cards and 6 blue cards giving 4+6 = 10 total.

The probability of picking not blue (aka picking red) is 4/10 because there are 4 red out of 10 total.

If we don't replace the card we picked, then we have 3 red left over out of 9 total. The probability of picking red again is 3/9

Multiply those fractions
(4/10)*(3/9) = (4*3)/(10*9) = 12/90

Now reduce
12/90 = (6*2)/(6*15) = 2/15
note how the GCF 6 cancels

Using a calculator, 
2/15 = 0.1333 approximately
The '3's after the decimal point go on forever.

The probability of both the cards not being blue would be:

-[tex]2/15 or 0.1333[/tex]

Given that,

Total cards = 10

No. of Red Cards = 4

No. of Blue cards = 6

The probability of not picking a blue card = 4/10     (∵ 4 are red)

As we know,

2 cards are picked at random,

In case,

The cards picked are not replaced, the possible chances of taking red would be = 3/9

Now,

The Probability of not picking a blue card(P) = no. of favorable outcomes/total outcomes

[tex](4/10)[/tex]× [tex](3/9)[/tex]

[tex]= (4[/tex]×[tex]3)/(10[/tex]×[tex]9)[/tex]

[tex]= 12/90[/tex]

By solving,

12/90

[tex]= (6[/tex]×[tex]2)/(6[/tex]×[tex]15)[/tex]

[tex]= 2/15[/tex]

Thus, 2/15 or 0.1333 is the correct answer.

Learn more about 'Probability' here:

brainly.com/question/11234923

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