Suppose that we roll a fair die until a 6 comes up.&nbsp;1) What is the probability that we roll the die n times?&nbsp;2) What is the expected number of times we roll the die?

Question

Suppose that we roll a fair die until a 6 comes up.&amp;nbsp;1) What is the probability that we roll the die n times?&amp;nbsp;2) What is the expected number of times we roll the die?

Wendy Boykin · Accepted Answer

1) If we roll a fair die, then we have 1 chance in 6 of rolling a six and 5 chances in 6 of not rolling a 6.&amp;nbsp;P(6)=16&amp;nbsp;P(Not&amp;nbsp;6)=56&amp;nbsp;We keep rolling the die until a 6 comes up.Let X be the number of rolls of the die.&amp;nbsp;When we roll the die n times, then the first&amp;nbsp;n−1&amp;nbsp;rolls cannot be a 6, while the n-th roll has to be a 6:&amp;nbsp;P(X=n)=(P(Not&amp;nbsp;6))n−1P(6)=(56)n−1×16&amp;nbsp;Note: The variable X has a distribution that is based on geometry.&amp;nbsp;p=16.&amp;nbsp;2) The expected value of a random variable X with a geometric distribution is the reciprocal of the constant probability of success&amp;nbsp;p=P(6)&amp;nbsp;E(x)=1p=1P(6)=116=6

Heather Fulton · Accepted Answer

1) We would have to roll&amp;nbsp;n−1&amp;nbsp;numbers that are not 6 followed by a 6, so the probability is&amp;nbsp;(56)n−1⋅(16)&amp;nbsp;2)&amp;nbsp;E(N)=∑n=1∞n⋅(56)n−1(16)=(16)(11−(56))2=6.&amp;nbsp;This is a nice answer since after 6 rolls we would expect to have rolled exactly one 6.

nick1337 · Accepted Answer

1. If we roll the die n times (assuming&amp;nbsp;n≥1&amp;nbsp;and the dice is 6-sided and fair), then we must roll n-1 &quot;not 6&quot; rolls followed by 1 &quot;6&quot; roll. The probability of that is:(56)n−1(16)12. We could roll the die any number of times from 1 to infinite. Consider a random variable R which is the number of rolls to roll a 6. We could determine the expected number in two ways:i. We could say that R is a geometric RV with a chance of success of&amp;nbsp;16, so&amp;nbsp;E(R)=1(16)=6ii. We could show i. explicitly, using the formula for expected value:E(R)=∑s∈SR(s)p(s)=∑i=1∞(i)((56)i−1(16)1)=16(∑i=1∞(i)((56)i−1)This summation takes the form∑i=1∞(i)(ri−1), where&amp;nbsp;|r|&amp;lt;1∑i=1∞(i)(ri−1)=∑i=1∞ddr(ri)=ddr∑i=1∞(ri)=ddr11−r&amp;nbsp;(since&amp;nbsp;|r|&amp;lt;1)=1(1−r)2Let&amp;nbsp;r=56(|r|=56&amp;lt;1.) Then,E(R)=16(1(1−56)2)⇒E(R)=6Either way, we get the same answer: 6 rolls.

alenahelenash · Accepted Answer

1) To find the probability of rolling the die n times until a 6 comes up, we can use geometric probability. The probability of rolling any number other than 6 in a single roll is 56, and since the rolls are independent, we can use the geometric probability formula.The probability of rolling the die n times until a 6 comes up is given by:P(n)=(56)n−1·162) To find the expected number of times we roll the die until a 6 comes up, we can use the concept of expected value. The expected value, denoted as E(X), is the sum of all possible values of a random variable X multiplied by their respective probabilities.In this case, the random variable X represents the number of rolls until a 6 comes up. We have already determined the probability of rolling the die n times, so we can use that to calculate the expected value.The expected number of times we roll the die is given by:E(X)=∑n=1∞n·P(n)Substituting the probability expression from part 1, we can write:E(X)=∑n=1∞n·(56)n−1·16To simplify this expression, we can differentiate the geometric series formula ∑n=0∞xn=11−x with respect to x:ddx(∑n=0∞xn)=ddx(11−x)This gives us:∑n=1∞nxn−1=1(1−x)2Now, differentiating both sides with respect to x again:∑n=1∞n(n−1)xn−2=2(1−x)3Multiplying both sides by x:∑n=1∞n(n−1)xn−1=2x(1−x)3Finally, substituting x=56, we get the simplified expression for the expected value:E(X)=2·56(1−56)3Simplifying further:E(X)=125Therefore, the expected number of times we roll the die until a 6 comes up is 125.

star233 · Accepted Answer

Answer:6Explanation:1) The probability of rolling the die exactly n times before obtaining a 6 can be calculated using geometric probability. In each roll, the probability of not rolling a 6 is 56 (since the die is fair). Therefore, the probability of rolling the die n times and obtaining a 6 on the nth roll is given by:P(X=n)=(56)n−1·16Here, X represents the number of rolls required to obtain a 6.2) The expected value or expected number of times we roll the die can be found by summing the product of each possible number of rolls and their respective probabilities. Mathematically, it is given by:E(X)=∑n=1∞n·P(X=n)To calculate the expected value, we can rewrite it as:E(X)=∑n=1∞n·(56)n−1·16To simplify the calculation, we can use the formula for the sum of an infinite geometric series:E(X)=16∑n=1∞n·(56)n−1The sum on the right side of the equation can be evaluated using the formula:∑n=1∞n·xn−1=1(1−x)2Substituting x=56, we have:E(X)=16·1(1−56)2Simplifying further, we get:E(X)=16·62=6Therefore, the expected number of times we roll the die before obtaining a 6 is 6.

Suppose that we roll a fair die until a 6

Answered question

Answer & Explanation

New Questions in High school probability