Question

In an experiment, two fair dice are thrown. (a) If we denote an outcome as the ordered pair (number of dots on the first die, number of dots on the second die), write down the sample space for the experiment. (So a roll of “1 dot” on the first die and a roll of “3 dots” on the second die would be the ordered pair (1, 3) in the sample space S.) You can think of the first die as red and the second die as green if that helps. (b) List the outcomes in the event “the sum of the dots on both dice is greater than or equal to 11”. (c) If both die are fair and independently thrown, what is the probability of the event “the sum of the dots on both dice is greater than or equal to 9”? (d) If both die are fair and independently thrown, what is the probability of the event “the sum of the dots on both dice is less than 7 and is not a ‘double’ ”? A ‘double’ is the same number of dots on both dice, e.g., (3, 3) or (6, 6).

Answer #1

a)

below is sample space

(1,1) | (1,2) | (1,3) | (1,4) | (1,5) | (1,6) |

(2,1) | (2,2) | (2,3) | (2,4) | (2,5) | (2,6) |

(3,1) | (3,2) | (3,3) | (3,4) | (3,5) | (3,6) |

(4,1) | (4,2) | (4,3) | (4,4) | (4,5) | (4,6) |

(5,1) | (5,2) | (5,3) | (5,4) | (5,5) | (5,6) |

(6,1) | (6,2) | (6,3) | (6,4) | (6,5) | (6,6) |

b)

outcomes for which sum is greater than or equal to 11 ={(5,6),(6,5),(6,6)}

c)total events =36

number of events sum is greater than or equal to 9 =N(sum 9)+N(sum 10)+N(sum 11)+N(sum 12)

=4+3+2+1=10

hence P( sum of the dots on both dice is greater than or equal to 9)=10/36=5/18

d)

number of events the sum of the dots on both dice is less than 7 and is not a ‘double’

=12

hence probability =12/36=1/3

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