If you roll a die 8 times, what is the probabiliy of getting both 1 and 2.
$P(1) = .3$
$ P(2) = .1$
$P(3) = .15$
$ P(4) = .15$
$P(5) = .15$
$P(6) = .15$
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I guess the easiest way to solve this is: 1-P(no 1)-P(no 2)+P(no 1 or 2) = $1-0.7^8-0.9^8+0.6^8$ We have to add +P(no 1 or 2) because otherwise we would be subtracting it twice. Ie. P(no 1) includes the case where we don't roll 1 or 2 and so does the P(no 2). Another way would be to use Markov chain with states S={no 1 or 2, only 1, only 2, 1 and 2}. Corresponding transition matrix is: $ D = \left[ \begin{array}{ccc} 0.6 & 0.3 & 0.1 & 0 \\ 0 & 0.9 & 0 & 0.1 \\ 0 & 0 & 0.7 & 0.3 \\ 0 & 0 & 0 & 1 \end{array} \right] $ Then calculate $D^8$ etc. |
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If you know about generating functions, you can do it as follows: Let $a=0.3, b= 0.1, c=1-a-b$, then the answer is: \begin{align} \int_0^r\int_0^s\left.\frac{d^2(ax + by + c)^8}{dxdy}dxdy\right|_{r=s=1} \end{align} But, really, for this problem Seth's suggestion is best. Try to break it into different cases whose probability you can easily calculate. (You can do it with three terms.) |
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=1- P(you don't roll a 1 or 2) |
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The probability of exactly a 1 once and a 2 once is 0.3(0.1)(0.6$^6$)=0.03(0.046656)=0.00139968 for each possible position of a 1 and a 2 to occur. So we multiply 0.03 by that number. The possiblilities are So the probability of exactly one 1 and one 2 is 56(0.00139968)=0.07838208 In addition to this we would add probability of two 1s and one two, one 1 and two 2s, two 1s and two 2s etc. |
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