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Tha 4

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Submitted By sharb
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THA 4

1) Betting game
You roll a die. If an odd number comes up, you lose. If you get a 6, you win $60. If it is an even number other than 6, you get to roll again. If you get a 6 the second time, you win $36. If not, you lose. (a) Construct a probability model for the amount you win at this game. Explain briefly how you obtain the probabilities associated with the different amounts of winning.
1,3,5=You lose
6first roll=you win $60
2,4=Roll again
6 (second roll)=You win $36

First roll | Second roll | P(x) | Winning amount | 1 | - | 1/6 | $0 | 3 | - | 1/6 | $0 | 5 | - | 1/6 | $0 | 6 | - | 1/6 | $60 | 2 | 1 | (1/6 * 1/6)= 1/36 | $0 | 2 | 2 | 1/36 | $0 | 2 | 3 | 1/36 | $0 | 2 | 4 | 1/36 | $0 | 2 | 5 | 1/36 | $0 | 2 | 6 | 1/36 | $36 | 4 | 1 | 1/36 | $0 | 4 | 2 | 1/36 | $0 | 4 | 3 | 1/36 | $0 | 4 | 4 | 1/36 | $0 | 4 | 5 | 1/36 | $0 | 4 | 6 | 1/36 | $36 |

Amount Probability Winnings | Probability | $0 | 316+10136=79=0.77=78% | $36 | 136+136=118=0.05=5% | $60 | 16=0.16=17% |

b) How much would you be willing to pay to play this game?

Outcome | Payout (x) | P(X)* | Win on first roll | $60 | 17% | Win on second roll | $36 | 5% | Lose | $0 | 78% |

E(x) = 60(0.17) + 36(0.05) + 0(0.78)
= 10.2 + 1.8 + 0
= 12
In order to make money from this game, statistically the bets cannot exceed over $12.

2) Chris Again! A true-false test consists of 10 items. (a) If Chris does not study at all and guesses each and every item in the test, describe the probability model for the number of correct guesses. We know that there’s a .5 chance of getting the question correct, also there’s a .5 chance of getting the question incorrect. Amount of correct question | Probability | (P)x | 1 | 0.5 | 0.5 | 2 | 0.5 * 0.5 = 0.5^2 | 0.25 | 3 | 0.5 * 0.5 * 0.5 = 0.5^3 | 0.125 | 4 | 0.5 * 0.5 * 0.5 * 0.5 = 0.5^4 | 0.0625 | 5 | 0.5^5 | 0.03125 | 6 | 0.5^6 | 0.015625 | 7 | 0.5^7 | 0.0078125 | 8 | 0.5^8 | 0.00390625 | 9 | 0.5^9 | 0.001953125 | 10 | 0.5^10 | 0.0009765625 |

(b) What is the probability that Chris gets 80% or more for the test? Using Binomial Experiment n=no. of trials π=prob. of success
X=no. of success

n=10 π= .5
X=≥8
using binomial table x | 0.05 | 0.10 | 0.20 | 0.25 | 0.30 | 0.40 | 0.50 | 0 | 0.599 | 0.349 | 0.107 | 0.056 | 0.028 | 0.006 | 0.001 | 1 | 0.315 | 0.387 | 0.268 | 0.188 | 0.121 | 0.040 | 0.010 | 2 | 0.075 | 0.194 | 0.302 | 0.282 | 0.233 | 0.121 | 0.044 | 3 | 0.010 | 0.057 | 0.201 | 0.250 | 0.267 | 0.215 | 0.117 | 4 | 0.001 | 0.011 | 0.088 | 0.146 | 0.200 | 0.251 | 0.205 | 5 | 0.000 | 0.001 | 0.026 | 0.058 | 0.103 | 0.201 | 0.246 | 6 | 0.000 | 0.000 | 0.006 | 0.016 | 0.037 | 0.111 | 0.205 | 7 | 0.000 | 0.000 | 0.001 | 0.003 | 0.009 | 0.042 | 0.117 | 8 | 0.000 | 0.000 | 0.000 | 0.000 | 0.001 | 0.011 | 0.044 | 9 | 0.000 | 0.000 | 0.000 | 0.000 | 0.000 | 0.002 | 0.010 | 10 | 0.000 | 0.000 | 0.000 | 0.000 | 0.000 | 0.000 | 0.001 | px≥8=pX=8+pX=9+pX=10
=0.044+0.010+0.001
= 0.055*100
=5.5%
Therefore the probability that Chris gets 80% or more for the test is 5.5%. (Note the binomial table is also rounded to the nearest hundredth decimal place).

Using formula 0 correct=0.510*1=0.009765625
1 correct=0.510*10!9!1!=0.009765625*10=0.009765625
2 correct=0.510*10!8!2!=0.009765625*45=0.043945313
3 correct=0.510*10!7!3!=0.009765625*120=0.1171875
4 correct=0.510*10!6!4!=0.205078125
5 correct=0.510*10!5!5!=0.24609375
6 correct=0.510*10!4!6!=0.205078125
7 correct=0.510*10!3!7!=0.1171875
8 correct=0.510*10!2!8!=0.43945313
9 correct=0.510*10!1!9!=0.009765625
10 correct=0.510*1=0.009765625

px≥8=pX=8+pX=9+pX=10
0.043945313 + 0.009765625 + 0.0009765625 = 0.0546875005 * 100
= 5.46%
The probability of Chris getting 80% or more on the test is 5.46%.

(c) If it is a 20 item true-false test, would you think it is easier or more difficulty for Chris to get 80% or more? Explain without performing any further calculation.
As seen above, the graph shows the distribution of the probability of the amount Chris got correct. As it can be seen ‘5’ is highest possibility with nearly .25, if the total amount of question was increase to 20 rather then 10, the distribution of the probability will increase and lower the highest probability. This will make it harder for Chris to achieve 80% or more.…...

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