I have this question which I can't figure out, maybe because the answer I got is not correct (trying to match the answer).

A six-sided die is thrown twice, but not a normal die. If X is the result of the die, then:

$$ P(X=1)=P(X=4)=0.2 $$

$$ P(X=2)=P(X=3)=P(X=6)=0.1$$

$$ P(X=5)=0.3$$

I am interested in the probability of getting an even number twice.

The answer I got is 0.06.

I can't figure out why... the explanation is:

$$ 0.04+2*0.01 $$

But it makes no sense to me.

  • $\begingroup$ You calculated the probability of getting an even number and then that same even number again. $\endgroup$ – AlexR Jun 6 '16 at 22:06

$$P(Even) = P(2)+P(4)+P(6)$$ $$~~~~~~= 0.1+0.2+0.1$$ $$ = 0.4~~~~~~~~~~~~~$$

P(Consecutive Even) equals the probability that role 1 and 2 are even. Since they are independent throws then you can multiply the 2 probabilities. 0.4*0.4 = 0.16

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