I need help figuring out some discrete probabilities.
$begingroup$
If I have 3 fair 6-sided dice I need to find the probability that 4 events occur.
First, the chance that I roll no 1's. I think the probability is $(5/6)^3$. Then the second event is all 1's, I think the probability is $(1/6)^6$. Next is exactly one 1, and finally exactly two 1's.
I have no idea how to do these last two. My first guess was that exactly one 1 would be $(1/6)*(5/6)^2$ and two 1's would be $(1/6)^2*(5/6)$, but when I added all four values together I got a sum of about 0.7, so I know it's not right.
probability
asked Jan 17 at 18:54
Matthew RileyMatthew Riley
896
$endgroup$
add a comment |
$begingroup$
If I have 3 fair 6-sided dice I need to find the probability that 4 events occur.
First, the chance that I roll no 1's. I think the probability is $(5/6)^3$. Then the second event is all 1's, I think the probability is $(1/6)^6$. Next is exactly one 1, and finally exactly two 1's.
I have no idea how to do these last two. My first guess was that exactly one 1 would be $(1/6)*(5/6)^2$ and two 1's would be $(1/6)^2*(5/6)$, but when I added all four values together I got a sum of about 0.7, so I know it's not right.
probability
asked Jan 17 at 18:54
Matthew RileyMatthew Riley
896
$endgroup$
add a comment |
$begingroup$
If I have 3 fair 6-sided dice I need to find the probability that 4 events occur.
First, the chance that I roll no 1's. I think the probability is $(5/6)^3$. Then the second event is all 1's, I think the probability is $(1/6)^6$. Next is exactly one 1, and finally exactly two 1's.
I have no idea how to do these last two. My first guess was that exactly one 1 would be $(1/6)*(5/6)^2$ and two 1's would be $(1/6)^2*(5/6)$, but when I added all four values together I got a sum of about 0.7, so I know it's not right.
probability
asked Jan 17 at 18:54
Matthew RileyMatthew Riley
896
$endgroup$
If I have 3 fair 6-sided dice I need to find the probability that 4 events occur.
First, the chance that I roll no 1's. I think the probability is $(5/6)^3$. Then the second event is all 1's, I think the probability is $(1/6)^6$. Next is exactly one 1, and finally exactly two 1's.
I have no idea how to do these last two. My first guess was that exactly one 1 would be $(1/6)*(5/6)^2$ and two 1's would be $(1/6)^2*(5/6)$, but when I added all four values together I got a sum of about 0.7, so I know it's not right.
probability
probability
asked Jan 17 at 18:54
Matthew RileyMatthew Riley
896
asked Jan 17 at 18:54
Matthew RileyMatthew Riley
896
asked Jan 17 at 18:54
Matthew RileyMatthew Riley
896
asked Jan 17 at 18:54
Matthew RileyMatthew Riley
896
asked Jan 17 at 18:54
Matthew RileyMatthew Riley
896
896
add a comment |
add a comment |
1 Answer
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$begingroup$
The first two are correct. You have the right idea for the second two, but you're a bit off.
For exactly one $1$, your probability of $frac{1}{6}(frac{5}{6})^{2}$ can be thought of as "roll a $1$ and then don't roll a $1$ for the next $2$ rolls". However the $1$ can occur on any of the three rolls, so this needs to be multiplied by $3$:
$$3cdotfrac{1}{6}cdotleft(frac{5}{6}right)^{2}$$
For exactly two $1$s, we'll use the same logic as above. There are $binom{3}{2}$ possible positions for the $1$s, and then a $(1/6)^{2}$ probability to roll them, and a $5/6$ probability to avoid them on the other rolls. So
$$binom{3}{2}left(frac{1}{6}right)^{2}frac{5}{6}$$
answered Jan 17 at 19:00
pwerthpwerth
3,233417
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1 Answer
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$begingroup$
The first two are correct. You have the right idea for the second two, but you're a bit off.
For exactly one $1$, your probability of $frac{1}{6}(frac{5}{6})^{2}$ can be thought of as "roll a $1$ and then don't roll a $1$ for the next $2$ rolls". However the $1$ can occur on any of the three rolls, so this needs to be multiplied by $3$:
$$3cdotfrac{1}{6}cdotleft(frac{5}{6}right)^{2}$$
For exactly two $1$s, we'll use the same logic as above. There are $binom{3}{2}$ possible positions for the $1$s, and then a $(1/6)^{2}$ probability to roll them, and a $5/6$ probability to avoid them on the other rolls. So
$$binom{3}{2}left(frac{1}{6}right)^{2}frac{5}{6}$$
answered Jan 17 at 19:00
pwerthpwerth
3,233417
$endgroup$
add a comment |
$begingroup$
The first two are correct. You have the right idea for the second two, but you're a bit off.
For exactly one $1$, your probability of $frac{1}{6}(frac{5}{6})^{2}$ can be thought of as "roll a $1$ and then don't roll a $1$ for the next $2$ rolls". However the $1$ can occur on any of the three rolls, so this needs to be multiplied by $3$:
$$3cdotfrac{1}{6}cdotleft(frac{5}{6}right)^{2}$$
For exactly two $1$s, we'll use the same logic as above. There are $binom{3}{2}$ possible positions for the $1$s, and then a $(1/6)^{2}$ probability to roll them, and a $5/6$ probability to avoid them on the other rolls. So
$$binom{3}{2}left(frac{1}{6}right)^{2}frac{5}{6}$$
answered Jan 17 at 19:00
pwerthpwerth
3,233417
$endgroup$
add a comment |
$begingroup$
The first two are correct. You have the right idea for the second two, but you're a bit off.
For exactly one $1$, your probability of $frac{1}{6}(frac{5}{6})^{2}$ can be thought of as "roll a $1$ and then don't roll a $1$ for the next $2$ rolls". However the $1$ can occur on any of the three rolls, so this needs to be multiplied by $3$:
$$3cdotfrac{1}{6}cdotleft(frac{5}{6}right)^{2}$$
For exactly two $1$s, we'll use the same logic as above. There are $binom{3}{2}$ possible positions for the $1$s, and then a $(1/6)^{2}$ probability to roll them, and a $5/6$ probability to avoid them on the other rolls. So
$$binom{3}{2}left(frac{1}{6}right)^{2}frac{5}{6}$$
answered Jan 17 at 19:00
pwerthpwerth
3,233417
$endgroup$
The first two are correct. You have the right idea for the second two, but you're a bit off.
For exactly one $1$, your probability of $frac{1}{6}(frac{5}{6})^{2}$ can be thought of as "roll a $1$ and then don't roll a $1$ for the next $2$ rolls". However the $1$ can occur on any of the three rolls, so this needs to be multiplied by $3$:
$$3cdotfrac{1}{6}cdotleft(frac{5}{6}right)^{2}$$
For exactly two $1$s, we'll use the same logic as above. There are $binom{3}{2}$ possible positions for the $1$s, and then a $(1/6)^{2}$ probability to roll them, and a $5/6$ probability to avoid them on the other rolls. So
$$binom{3}{2}left(frac{1}{6}right)^{2}frac{5}{6}$$
answered Jan 17 at 19:00
pwerthpwerth
3,233417
answered Jan 17 at 19:00
pwerthpwerth
3,233417
answered Jan 17 at 19:00
pwerthpwerth
3,233417
answered Jan 17 at 19:00
pwerthpwerth
3,233417
3,233417
add a comment |
add a comment |
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