Statistics

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Question 1

For the first trial the probability of winning can be presented as;

Possible out come | The probability of winning | fraction gained |

1 | 5/6 | |

2 | 5/6 | |

3 | 5/6 | |

4 | 5/6 | |

5 | 5/6 | |

6 | 56 |

Tacking the frequency of winning to be F, the distribution function can be given by

The expected value sequence therefore is,

The gives then can be given by;

The odd is not fair since the probability of gaining is very low.

Question 2

- situation 1

The probability that the first marble is red and the second one is green is given by;

This is because the marbles remains two in number in the second pick since there is no replacement after the first picking.

Situation 2

The probability that the first marble is one and the second one is green where there is replacement after picking is given by;

The probability of picking green in the second attempt remains a third since the total number goes back to three due to replacement.

- Probability of C

- situation 1

Event A is given by;

Event B is given by;

The probability that event A and B occurring is given by;

Event A× Event B

- situation 2- Drawing with replacement:

Event A;

Event B;

The probability that event A and B occurring is given by;

Event A× Event B

The Probability of E

Situation 1

The probability that A occurs is given by;

The probability that event B occur;

The probability that Event A or B occurs is given by;

Probability of Event A occurring + The probability of Event B occurring

Situation 2;

Event A;

Event B;

The probability of event A or B occurring can be given by;

Event A + Event B

- the probability that A does not occur:

Situation 1

The probability that Event A does not occur can be given by;

1– The probability of eventA occurring.

Situation 2

Likewise the probability that event A does not occur is given by;

Question 3;

- the probability that there will be at least one tail;

1– The probability of obtaining five tails.

- the probability that there will be at most one head;

the probability of obtaining one head + the probability of obtaining all tails

P (one head) =

P(all tails)=

The probability of obtaining at least one head is therefore;

- the probability of obtaining one head;

Let n be the number of trials which is equal to 5 and R be the number of success which is 1,

The number of failure would be:

From a single toss,

P(success)=p(failure)=0.5

The probability of obtaining 1 head is therefore given by;

P (success)^{R}×p(success)^{n-R} which is 0.5^{1}×0.5^{4}=0.03125;

The possible ways of obtaining a head is given by polynomial expansion:

_{N}C_{R} = _{5}C_{4} =5

The probability of obtaining one head from five tosses is therefore;

5(0.03125) =

It more suitable to use binomial probability formula since the data is widely distributed and considering individual situation will be tiresome. Besides binomial binomial probability formula works best in this situation since the population is remarkably larger that the sample size

Question 4

This is because the ball has an equal probability of landing any pocket (black or red), even or odd, small digits or large digits. The probability of winning a simple bet and their elements is as shown;

Simple bet | Probability odds | Pay out ($) |

Straight up | 35 to 1 | |

Low or high bet | 1 to 1 | |

Even or odd bet | 1 to 1 | |

Color bet | 1 to 1 | |

Dozen bet | 2 to 1 | |

Colum bet | 2 to 1 | |

Line bet | 5 to 1 | |

Corner bet | 8 to 1 | |

Street bet | 11 to 1 | |

Split bet | 17 to 1 | |

Straight up | 35 to 1 |

The chance of winning a bet is determined by the position on which a bet is placed and the number of bet attempted and its given by the product ofthe set off all roulette number by the set of group of numbers from the set of all roulette number made by the unique placement which are allowed for betting. The average law states that the chances of a ball falling into a slot are one out of the 38 slots available for the entire slot and this is very independent of the previous bet. Sampling a long range of independent possible chances and averaging them may be used to obtain a stable prediction of the possible outcome.

References

Freedman, D., Pisani, R., & Purves, R. (2008). *Statistics*. New York: Norton.

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