There are 3 red and 4 blue marbles in a bag. At random, one marble is

1.	There are 3 red and 4 blue marbles in a bag. At random, one marble is drawn. The probability that the marble drawn is blue :(A) 1(B) \(\frac { 3 }{ 7 }\)(C) \(\frac { 4 }{ 7 }\)(D) \(\frac { 2 }{ 7 }\)
Answer» Answer is (C) \(\frac { 4 }{ 7 }\) Total number of marbles in the bag = 3 + 4 = 7 ∴ The number of all possible outcomes = 7 The number of favourable outcomes of getting blue marbles = 4 ∴ Required probability = \(\frac { 4 }{ 7 }\)

There are 3 red and 4 blue marbles in a bag. At random, one marble is drawn. The probability that the marble drawn is blue :(A) 1(B) \(\frac { 3 }{ 7 }\)(C) \(\frac { 4 }{ 7 }\)(D) \(\frac { 2 }{ 7 }\)

Answer»

Answer is (C) \(\frac { 4 }{ 7 }\)

Total number of marbles in the bag = 3 + 4 = 7

∴ The number of all possible outcomes = 7

The number of favourable outcomes of getting blue marbles = 4

∴ Required probability = \(\frac { 4 }{ 7 }\)

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There are 3 red and 4 blue marbles in a bag. At random, one marble is drawn. The probability that the marble drawn is blue :(A) 1(B) \(\frac { 3 }{ 7 }\)(C) \(\frac { 4 }{ 7 }\)(D) \(\frac { 2 }{ 7 }\)

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