Samuel Pepys was a great diarist of the English language He sought Sir Issac New

Question

Samuel Pepys was a great diarist of the English language He sought Sir Issac Newton's advice in 1693 on a matter that related to gambling. Is it more likely to get at least one 6 when 6 dice are rotted, at least two 6's when 12 dice are rolled or at least 3 6's when 18 dice are rolled? Newton was able to convince Pepys that the former was most likely. Compute each of these probabilities to confirm Newton. Assume the die are fair. See below for more. For each of the 3 scenarios what are the n trials? One a single trial what is considered a success and a failure? What is the probability of a success? Let X= number of success in n trials be your binomial random variable and find for example P(X greaterthanequal l)or P(X greaterthanequal 2) Or P(X greaterthanequal 3) depending on which scenario.

Explanation / Answer

there are three scenarios

getting at least one 6 when 6 dice are rolled

getting at least 2 6's when 12 dice are rolled

getting at least 3 6's when 18 dice are rolled

1. so for the first scenario number of trials=n=6

for the second scenario number of trials=n=12

for the third scenario number of trials=n=18

2. on a single trial getting a 6 is considered as success and not getting a six is considered as failure

3. since the assumption is that the die is fair.

so the probability of success=P[getting a six]=1/6 [answer]

4. here for the first scenario

X~Bin(6,1/6)

so pmf is P[X=x]=6Cx(1/6)x(1-1/6)6-x      x=0,1,2,....,6

so P[X>=1]=1-P[X=0]=1-6C0(1/6)0(1-1/6)6-0=0.665 [answer]

for the second scenario X~Bin(12,1/6)

so pmf is P[X=x]=12Cx(1/6)x(1-1/6)12-x      x=0,1,2,....,12

so P[X>=2]=1-P[X=0]-P[X=1]=1-12C0(1/6)0(1-1/6)12-0-12C1(1/6)1(1-1/6)12-1=1-0.112-0.269=0.619

for the third scenario X~Bin(18,1/6)

so pmf of X is P[X=x]=18Cx(1/6)x(1-1/6)18-x          x=0,1,2,.......,18

so P[X>=3]=1-P[X=0]-P[X=1]-P[X=2]=1-18C0(1/6)0(1-1/6)18-0-18C1(1/6)1(1-1/6)18-1-18C2(1/6)2(1-1/6)18-2

                      =1-0.038-0.135-0.230=0.597 [answer]

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