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We throw a fair 6-sided die independently four times and let $X$ denote the minimal value rolled.

  1. What is the probability that $X \ge 4$?

  2. Compute the PMF of $X$.

  3. Determine the mean and variance of $X$.

My attempt:

  1. $(1/2)^4$ because that means each of the $4$ rolls, you either get a $4, 5$ or $6$.

  2. for $X=1: (1/6)(6/6)(6/6)(6/6)$
    for $X=2: (1/6)(5/6)(5/6)(5/6)$
    for $X=3: (1/6)(4/6)(4/6)(4/6)$
    for $X=4: (1/6)(3/6)(3/6)(3/6)$
    for $X=5: (1/6)(2/6)(2/6)(2/6)$
    for $X=6: (1/6)(1/6)(1/6)(1/6)$

  3. I can calculate this once I know I did the PMF correctly

Did I do (1) and (2) correctly?

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  • $\begingroup$ For (1) you need $\left(\frac{1}{2}\right)^4$ $\endgroup$ – Daniel Mathias Feb 7 at 19:29
  • $\begingroup$ ad a) But you have to roll 4 times. So $P(X=4)=\left(\frac{1}{2}\right)^4$ And $P(X=5)=\left(\frac{1}{3}\right)^4$ And al last $P(X=6)=\left(\frac{1}{6}\right)^4$. To get $P(X\geq 4)$ you have to sum up the probabilities. $\endgroup$ – callculus Feb 7 at 19:31
  • $\begingroup$ @calculus that should be $P(X\ge4)$ and $P(X\ge5)$ $\endgroup$ – Daniel Mathias Feb 7 at 19:34
  • $\begingroup$ @DanielMathias No, since $X$ denote the minimal value rolled $\endgroup$ – callculus Feb 7 at 19:35
  • $\begingroup$ @callculus for $X=4$, one of the rolls must be 4. But $\left(\frac{1}{2}\right)^4$ includes the events that all rolls are greater than 4. $\endgroup$ – Daniel Mathias Feb 7 at 19:38
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We throw a fair 6-sided die independently four times and let $X$ denote the minimal value rolled.

What is the probability that $X \ge 4$?

For each roll, the value must be greater than three. This event has probability $\frac{3}{6}=\frac{1}{2}$. As this must occur on each of $4$ rolls, we have: $$P(X\ge4)=\left(\frac{1}{2}\right)^4=\frac{1}{16}$$

Compute the PMF of $X$.

$$P(X=1)=P(X>0)-P(X>1)=1-\left(\frac{5}{6}\right)^4=1-\frac{625}{1296}=\frac{671}{1296}$$

$$P(X=2)=P(X>1)-P(X>2)=\left(\frac{5}{6}\right)^4-\left(\frac{4}{6}\right)^4=\frac{625}{1296}-\frac{256}{1296}=\frac{369}{1296}$$

$$P(X=3)=P(X>2)-P(X>3)=\left(\frac{4}{6}\right)^4-\left(\frac{3}{6}\right)^4=\frac{256}{1296}-\frac{81}{1296}=\frac{175}{1296}$$

$$P(X=4)=P(X>3)-P(X>4)=\left(\frac{3}{6}\right)^4-\left(\frac{2}{6}\right)^4=\frac{81}{1296}-\frac{16}{1296}=\frac{65}{1296}$$

$$P(X=5)=P(X>4)-P(X>5)=\left(\frac{2}{6}\right)^4-\left(\frac{1}{6}\right)^4=\frac{16}{1296}-\frac{1}{1296}=\frac{15}{1296}$$

$$P(X=6)=P(X>5)-P(X>6)=\left(\frac{1}{6}\right)^4-0=\frac{1}{1296}$$

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1) is now correct after your edit based on Daniel's comment.

2) is incorrect since a) it doesn't sum to 1 and b) $X = 4, 5, 6$ doesn't sum to what you got in 1).

My approach to this question is to break it down into a bunch of simpler problems as such:

For $P(X=1)$, let's consider the $4$ dice as dice $1, 2, 3, 4$. There are $4$ possibilities, the result can have $1,2,3,$ or $4$ ones in it.

  • $1$ one: ${4 \choose 1} (\frac{1}{6})^1(\frac{5}{6})^3$
  • $2$ ones: ${4 \choose 2} (\frac{1}{6})^2(\frac{5}{6})^2$
  • $3$ ones: ${4 \choose 3} (\frac{1}{6})^3(\frac{5}{6})^1$
  • $4$ ones: ${4 \choose 4} (\frac{1}{6})^4(\frac{5}{6})^0 = (\frac{1}{6})^4$

The sum of the above will give you the $P(X=1)$.

Similarly, you can solve for the rest. Make sure that the end results sums to $1$ and $P(X=4) + P(X=5) + P(X=6) = (\frac{1}{2})^4$.

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