A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?

**Options:**

A. | \(\frac{10}{21}\) |

B. | \(\frac{11}{21}\) |

C. | \(\frac{2}{7}\) |

D. | \(\frac{5}{7}\) |

**Answer: Option A**

Total number of balls = (2 + 3 + 2) = 7.

Let S be the sample space

Then, *n*(S) = Number of ways of drawing 2 balls out of 7

= ^{7}C_{2}

= \(\frac{(7\times6)}{(2\times1)}\)

=21.

Let E = Event of drawing 2 balls, none of which is blue.

So, *n*(E) = = Number of ways of drawing 2 balls out of (2 + 3) balls.

= ^{5}C_{2}

_{= \(\frac{(5\times4)}{(2\times1)}\)}

_{=10.}

So, \(P(E) = \frac{n(E)}{n(S)}=\frac{10}{21}\)

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