A) 21/46 | B) 1/5 |

C) 3/25 | D) 1/50 |

Explanation:

Let , S - sample space E - event of selecting 1 girl and 2 boys.

Then, n(S) = Number ways of selecting 3 students out of 25

= $25{C}_{3}$

= 2300.

n(E) = $10{C}_{1}\times 15{C}_{2}$ = 1050.

P(E) = n(E)/n(s) = 1050/2300 = 21/46

A) 21/46 | B) 1/5 |

C) 3/25 | D) 1/50 |

Explanation:

Let S be the sample space and E be the event of selecting 1 girl and 2 boys.

Then, n(S) = Number ways of selecting 3 students out of 25

= $25{C}_{3}$ = 2300.

n(E)= $10{C}_{1}*15{C}_{2}$ = 1050.

P(E) = n(E)/n(s) = 1050/2300 = 21/46

A) 3/4 | B) 3/8 |

C) 5/16 | D) 2/7 |

Explanation:

In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36.

Then, E= {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3,4),(3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1),6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

n(E) = 27.

P(E) = n(E)/n(S) = 27/36 = 3/4.

A) 3/4 | B) 7/8 |

C) 1/2 | D) 1/4 |

Explanation:

Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}

Let E = event of getting at most two heads.

Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}.

P(E) =n(E)/n(S)=7/8.

A) 1/2 | B) 3/4 |

C) 1/9 | D) 2/9 |

Explanation:

In two throws of a die, n(S) = (6 x 6) = 36.

Let E = event of getting a sum ={(3, 6), (4, 5), (5, 4), (6, 3)}.

P(E) =n(E)/n(S)=4/36=1/9.

A) 10/21 | B) 11/21 |

C) 1/2 | D) 2/7 |

Explanation:

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 =$7{C}_{2}$ = 21

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

n(E) = Number of ways of drawing 2 balls out of (2 + 3) balls =$5{C}_{2}$ = 10

Therefore, P(E) = n(E)/n(S) = 10/ 21.

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

C) 8/21 | D) 7/21 |

Explanation:

Total number of balls = (8 + 7 + 6) = 21.

Let E = event that the ball drawn is neither red nor green

= event that the ball drawn is blue.

n(E) = 7.

P(E) = n(E)/n(S) = 7/21 = 1/3.

A) 1/2 | B) 3/5 |

C) 9/20 | D) 8/15 |

Explanation:

Here, S = {1, 2, 3, 4, ...., 19, 20}.

Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.

P(E) = n(E)/n(S) = 9/20.