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An urn contains 7 red and 4 blue balls. Two balls are drawn with repla

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Bunuel
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An urn contains 7 red and 4 blue balls. Two balls are drawn with repla [#permalink] New post   26 Feb 2021, 10:30
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A
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E

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25% (medium)

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83% (01:30) correct 17% (01:50) wrong based on 30 sessions
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An urn contains 7 red and 4 blue balls. Two balls are drawn with replacement. What is the probability that the balls are both the same colour ?

A. 40/121
B. 21/55
C. 41/121
D. 56/121
E. 65/121
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E
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ankitgoswami
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Re: An urn contains 7 red and 4 blue balls. Two balls are drawn with repla [#permalink] New post   26 Feb 2021, 10:39
I think its E

Probability of both being red = 7/11 * 7/11 = 49/121

Probability of both being blue = 4/11 * 4/11 = 16/121

So total prob = (49+16)/121 = 65/121

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An urn contains 7 red and 4 blue balls. Two balls are drawn with repla [#permalink] New post   26 Feb 2021, 20:52
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Bunuel wrote:
An urn contains 7 red and 4 blue balls. Two balls are drawn with replacement. What is the probability that the balls are both the same colour ?

A. 40/121
B. 21/55
C. 41/121
D. 56/121
E. 65/121


Total Balls = 7+4 = 11
Red = 7
Blue = 4

Probability of a pick to be a red ball \(= \frac{7}{11}\)
Probability of second pick to be a red ball \(= \frac{7}{11}\)
Probablity of both to be Red balls = \(= \frac{7}{11}*\frac{7}{11} = \frac{49}{121}\)

Probability of a pick to be a Blue ball \(= \frac{4}{11}\)
Probability of second pick to be a Blue ball \(= \frac{4}{11}\)
Probablity of both to be Blue balls = \(= \frac{4}{11}*\frac{4}{11} = \frac{16}{121}\)

Probability fo Both to be same colour \(= \frac{49}{121}\)+\(\frac{16}{121}\) = \(\frac{65}{121}\)

ANswer: Option E

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Re: An urn contains 7 red and 4 blue balls. Two balls are drawn with repla [#permalink] New post   28 Feb 2021, 07:45
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Given
    • An urn contains 7 red and 4 blue balls.
    • Two balls are drawn with replacement.

To find

    • The probability that both the balls are of the same color

Approach
    • Probability of picking both the balls of same color= \(\frac{Total\ cases\ of\ picking\ (Both\ red\ balls\ +\ Both\ blue\ balls)\ }{ Total\ cases\ of\ picking\ 2\ balls\ with\ replacement}\)
      o Total cases of picking (Both red balls + Both blue balls) = 7 × 7 + 4 × 4
         49 + 16= 65
      o Total cases of picking 2 balls with replacement = 11 × 11 =121
    • • Probability of picking both the balls of same color = \(\frac{65 }{ 121}\)

Hence, the correct answer is option E.
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Re: An urn contains 7 red and 4 blue balls. Two balls are drawn with repla [#permalink] New post   28 Feb 2021, 07:57
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Probability choosing a red ball = 7/11
Probability choosing a blue ball = 4/11
Since, balls are drawn with replacement, hence for the second ball, the probability of choosing each ball would remain same. Therefore, probablity of choosing two balls of same colour = 7/11*7/11 + 4/11*4/11 = 65/121
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Re: An urn contains 7 red and 4 blue balls. Two balls are drawn with repla [#permalink]
28 Feb 2021, 07:57
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