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S1 Question

Got a question here.

Which is more likely; That you get at least one 6 when you throw a die 6 times or that you get at least two 6s when you throw it twelve times?

I've been doing these questions, but reading the question again, this time it's different. It says "at least" and the questions I usually do, they ask me for the number "exactly" if you get what I mean.

I thought of doing 6C1 x 1/6^1 x 5/6^5 but it turns out to be wrong.

Any help on this would be appreciated. Thanks.
X = Number of sixes in the first situation
Y = Number of sixes in the second situation

The numbers you want are:
P(X1)=1P(X=0)\mathrm{P(X\geq 1)}=1- \mathrm{P(X=0)}
P(Y2)=1(P(Y=0)+P(Y=1))\mathrm{P(Y\geq 2)}=1- \mathrm{(P(Y=0) + P(Y=1))}

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