Consider the following frequency table of second year Engg students taking various subjects:

Subject\Student | Male | Female |
---|---|---|

Calculus | 5 | 25 |

History | 5 | 30 |

Physics | 15 | 6 |

1. What is the probability that a 2nd Year Engg. student chosen is taking Calculus or is a Male?

2. What is the probability that a 2nd Year Enng. student is a Female or is taking History?

3. What is the probability that 2nd Year Engg. student is a female taking History?

4. What is the probability that a 2nd Year Engg. student is not taking Caculus?

Answers.

First form the full table with column and row sums.

Subject\Student | Male | Female | Sum | |
---|---|---|---|---|

Calculus | 15 | 25 | 40 | |

History | 5 | 30 | 35 | |

Physics | 15 | 6 | 21 | |

Total | 35 | 61 | 96 |

1.Use the fact that $$P(A\cup B) = P(A) + P(B) - P(A\capB)$$

The answer is then $$P(Calulus \cup Male) = P(Calculus) + P(Male) - P(Calculus \cap Male)$$

or $$\frac{40 + 35 - 15}{96}= \frac{60}{96}= 0.625$$

2. Similar to #1,$$P(History \cup Female) = P(Calculus) + P(Female) - P(History \cap Female)$$

Solving, $$\frac{61+35-30}{96}=\frac{66}{96}=\frac{11}{96}= 0.114583$$.

3. This is the simplest problem. From the above table, it is $$\frac{30}{96} = 0.3125$$

4. The complement is $$1 - P(Calculus)$$. Therefore $$1- \frac{40}{96} = \frac{54}{96} = 0.5625$$

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