Probability 47373

We were tasked with throwing the dice until we hit the "six."

a) Find the average number of throws we will have to make to complete the task.

b) How many times do we have to roll the dice so that the probability of falling at least one "six" is at least 90%

Correct answer:

a =  6
s =  13

Step-by-step explanation:

p1=610.1667 a=1/p1=1/61=1:61=1 16=11 6=16=6
pn>0.9 p2=p1+(1p1)1 p1=0.1667+(10.1667)1 0.1667=36110.3056 p3=p2+(1p1)2 p1=0.3056+(10.1667)2 0.1667=216910.4213 p4=p3+(1p1)3 p1=0.4213+(10.1667)3 0.1667=12966710.5177 p5=p4+(1p1)4 p1=0.5177+(10.1667)4 0.16670.5981 p6=p5+(1p1)5 p1=0.5981+(10.1667)5 0.16670.6651 p7=p6+(1p1)6 p1=0.6651+(10.1667)6 0.16670.7209 p8=p7+(1p1)7 p1=0.7209+(10.1667)7 0.16670.7674 p9=p8+(1p1)8 p1=0.7674+(10.1667)8 0.16670.8062 p10=p9+(1p1)9 p1=0.8062+(10.1667)9 0.16670.8385 p11=p10+(1p1)10 p1=0.8385+(10.1667)10 0.16670.8654 p12=p11+(1p1)11 p1=0.8654+(10.1667)11 0.16670.8878 p13=p12+(1p1)12 p1=0.8878+(10.1667)12 0.16670.9065  p11>0.90  s=13



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