Fair die tossing problems

[lhuyvn]

Hi All,

With the given problem below, any suggestion?

A fair die is tossed 360 times. The probability that a six comes up on 70 or more of the tosses is

A. greater than 0.50
B.between 0.16 and 0.50
C.between 0.02 and 0.16
D.between 0.01 and 0.02
E. less than 0.01

 

[HallsofIvy]

With 360 tosses it probably best to use the normal approximation to the binomial distribution. The probability of a 6 on one roll is 1/6 so the expected number of 6s in 360 rolls is (1/6)(360)= 60. The standard deviation is sqrt((1/6)(5/6)(360= sqrt(50)= 5sqrt(2). Since the number of 6s must be an integer while the normal variable is continuous, use the "half-integer" correction: instead of "more than 70" use "more than 69.5". The standard variable would be (69.5- 60)/(5sqrt(2)). Use a table of the standard normal distribution to find the probability that z is larger than that.

 

[jcsd]

Surely you can use the series:

$$\frac{1}{6^{360}}\sum^{290}_{n=0} 5^n = \frac{1-5^{291}}{6^{360}(1-5)}$$

 

[jcsd]

my answer seems ridculously low though and it doesn't equal unity where it should

edited to add I see the mistake now I've assumed a specific order so the series doesn't work.

 

[NateTG]

Well, you could take the binomial distribution approach. If you take the sum of the first 290 terms of:
$$(\frac{1}{6} + \frac{1}{6})^{360})$$
Then you get an exact answer.

 

[Gokul43201]

Surely you can use the series:

$$\frac{1}{6^{360}}\sum^{290}_{n=0} 5^n = \frac{1-5^{291}}{6^{360}(1-5)}$$

I think what you mean is :

$$1- \frac {1} {6^{360}}\sum^{70}_{n=0} 5^{360-n}$$

Edited to add : this makes it look like a near certainty (~ 1 - 10^-29), something's wrong !

 

[jcsd]

I think what you mean is :

$$1- \frac {1} {6^{360}}\sum^{70}_{n=0} 5^{360-n}$$

Edited to add : this makes it look like a near certainty (~ 1 - 10^-29), something's wrong !

I'm pretty sure the original equation was sound, but it needs the following modification to hold true:

$$\frac{1}{6^{360}}\sum^{290}_{n=0} ^{360}P_{n}5^n$$