A) 1/2 | B) 7/15 |

C) 8/15 | D) 1/9 |

Explanation:

Let S be the sample space

Then n(S) = no of ways of drawing 2 balls out of (6+4) =$10{C}_{2}$ 10 =$\frac{10*9}{2*1}$ =45

Let E = event of getting both balls of same colour

Then,n(E) = no of ways (2 balls out of six) or (2 balls out of 4)

=$6{C}_{2}+4{C}_{2}$ = $\frac{6*5}{2*1}+\frac{4*3}{2*1}$= 15+6 = 21

Therefore, P(E) = n(E)/n(S) = 21/45 = 7/15

A) 16 | B) 12 |

C) 10 | D) 14 |

Explanation:

Total balls = 40

Red balls = 18

Let green balls are x

Then, (18/40) × (

A) 3/5 | B) 1/5 |

C) 2/5 | D) 4/5 |

Explanation:

A) 11/36 | B) 25/36 |

C) 5/6 | D) 1/6 |

Explanation:

We first calculate the probability of getting an even number on one and a multiple of 3 on other,Here, n(s) = 6x6 = 36 and

E = (2,3) (2,6) (4,3) (4,6) (6,3) (6,6) (3,2) (3,4)(3,6) (6,2)(6,4)

n(E) = 11P(E) = 11/36Required probability = 1-11/36 = 25/36

A) 1/6 | B) 2/3 |

C) 14/33 | D) 5/6 |

Explanation:

Possible outcomes = (RY, YP, PR)

2C1 4C1 + 4C1 6C1 + 6C1 2C1

Required probability = (2C1 4C1 + 4C1 6C1 + 6C1 2C1)/(12C2)

= 8 + 24 + 12/66

= 44/66

= 2/3.

A) 2 | B) 1/2 |

C) 1/4 | D) 1 |

Explanation:

Probability is Number of favourable outcomes by total outcomes

Total comes = 2 (Divisible or not)

Favourable = 1 (Divisible)

Hence, **probability = 1/2.**

A) 0.4987 | B) 0.5 |

C) 0.9987 | D) 0.1217 |

Explanation:

Probability between z = 0 and z = 3.01 is given by

P(0<z<3.01) = P(z<3.01) - P(z<0)

Reading from the z-table, we have

P(z<0) = 0.5

P(z<3.01) = 0.9987

Hence, P(0<z<3.01) = 0.9987 - 0.5 = 0.4987.

A) 1 | B) 21/26 |

C) 5/26 | D) 5 |

Explanation:

We know that,

Total number of balls n(S) = 26

Number of vowels n(E) = 5

Hence, required probability = n(E)/n(S) = **5/26.**