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There are two boxes. Box A contains 6 red balls and 4 blue balls and B

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Bunuel
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There are two boxes. Box A contains 6 red balls and 4 blue balls and B [#permalink] New post   21 Sep 2020, 04:45
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There are two boxes. Box A contains 6 red balls and 4 blue balls and Box B contains 4 red balls and 8 blue balls. A die is rolled, if the number is less than 3, a ball is selected from box A. If the result is 3 or more, a ball is selected from Box B. What is the probability that the ball will be red and selected from Box B?

A. 1/9
B. 2/9
C. 19/45
D. 26/45
E. 7/9


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yashikaaggarwal
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There are two boxes. Box A contains 6 red balls and 4 blue balls and B [#permalink] New post   21 Sep 2020, 04:59
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Box A contains 6 red balls and 4 blue balls,
Probability of red in Box A = 6/10

and Box B contains 4 red balls and 8 blue balls.
Probability of red in Box B = 4/12

A die is rolled, if the number is less than 3, a ball is selected from box A.
Probability of getting 1,2 on dice = 2/6 = 1/3

If the result is 3 or more, a ball is selected from Box B.
Probability of getting 3-6 on dice = 4/6 = 2/3

What is the probability that the ball will be red and selected from Box B = 4/12*2/3 = 2/9

Answer is B
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Re: There are two boxes. Box A contains 6 red balls and 4 blue balls and B [#permalink] New post   26 Sep 2020, 17:44
we have to satify two condition.

first the dice must be greater than 3 or more so the probiltiy of dice giving a number 3 or more = 4/6

second condition is picking a red ball,as B contains 4 red balls and 8 blue balls. Probability of picking red in Box B = 4/12

to get the probility of the conditions mention in the above question we need get both the condition so we have to multiply those probalities = 4/6* 4/12 =2/9
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Re: There are two boxes. Box A contains 6 red balls and 4 blue balls and B [#permalink] New post   28 Oct 2020, 12:29
Probability for Red ball (Box B):
Red = \(\frac{4}{12}\) = 1/3
Probability that \(die\geq{3}\) = \(\frac{4}{6}\) = \(\frac{2}{3}\)
Probability of drawing a red ball from box B = \(\frac{2}{3}\) * \(\frac{1}{3}\) = \(\frac{2}{9}\)
Answer = B
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Re: There are two boxes. Box A contains 6 red balls and 4 blue balls and B [#permalink] New post   05 Nov 2020, 17:09
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Bunuel wrote:
There are two boxes. Box A contains 6 red balls and 4 blue balls and Box B contains 4 red balls and 8 blue balls. A die is rolled, if the number is less than 3, a ball is selected from box A. If the result is 3 or more, a ball is selected from Box B. What is the probability that the ball will be red and selected from Box B?

A. 1/9
B. 2/9
C. 19/45
D. 26/45
E. 7/9



Solution:

The probability of selecting box B (i.e., rolling a number greater than or equal to 3 on a die) is 4/6 = 2/3. The probability of selecting a red ball from box B is 4/12 = 1/3. Therefore, the probability of choosing a red ball from box B is 2/3 x 1/3 = 2/9.

Answer: B
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Re: There are two boxes. Box A contains 6 red balls and 4 blue balls and B [#permalink]
05 Nov 2020, 17:09
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