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Suppose you have $6$ fair weighted dice with all faces having $1$ to $6$. since the outcome is mutually independent and are disjointed sets, the probability of getting at least one six is $P(a)=6\cdot(1/6)$ by law of addition but if now there are $7$ dices so $P(a)$ is $7\cdot(1/6)$,how?

Now a follow up question -what is the probability of $6$ dices having Same sides except $6$ which is $P(b)=5(\frac16)^6$

But we know $P(a)$, probabability of getting at least $1$ six in $6$ dice is $1$, so this means $P(b)=0$ since one side must have six but earlier we exclude it.

Kenta S
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Piyush Choudhury
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  • I think you're confusing the probability of getting at least one six, with the expected number of sixes. No matter how many dice you roll, there is a nonzero probability of not rolling any sixes. – Rivers McForge Aug 07 '20 at 18:17

1 Answers1

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The probability of having at least one 6 is $1-(\tfrac{5}{6})^6$.

The events are not disjoint: you can have 6 in dice #1 and in dice #2 at the same time.

YJT
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