Question

You have a six-sided die that you roll once. Let RiR_{i} denote the event that the roll is i. Let GjG_{j} denote the event that the roll is greater than j. Let E denote the event that the roll of the die is even-numbered. (a) What is P[R3G1]\mathrm{P}\left[R_{3} | G_{1}\right], the conditional probability that 3 is rolled given that the roll is greater than 1? (b) What is the conditional probability that 6 is rolled given that the roll is greater than 3? (c) What is P[G3E]\mathrm{P}\left[G_{3} | E\right], the conditional probability that the roll is greater than 3 given that the roll is even? (d) Given that the roll is greater than 3, what is the conditional probability that the roll is even?

Solution

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Sample Space for rolling a die, S = {1, 2, 3, 4, 5, 6}

Since, all the outcomes are equally likely, P(rolling any number) = 16\dfrac{1}{6}

(a)\textbf{(a)}

As we can see, G1G_{1} is the sample space in this case.

G1G_{1} = {2, 3, 4, 5, 6}

Hence, P(G1G_{1}) = P(R2R_{2}) + P(R3R_{3}) + P(R4R_{4}) + P(R5R_{5}) + P(R6R_{6})

Hence, P(G1G_{1}) = 16+16+16+16+16=56\dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{5}{6}

and R3R_{3} = {3}

Hence, G1R3G_{1} \cap R_{3} = {3} = R3R_{3}

Hence, P(G1R3G_{1} \cap R_{3}) = P(R3R_{3}) = 16\dfrac{1}{6}

Hence, the conditional probability that 3 is rolled given that the roll is greater than 1 i.e.

P(R3G1R_{3} | G_{1}) = P(G1R3)P(G1)\dfrac{P(G_{1} \cap R_{3})}{P(G_{1})} = 1656\dfrac{\frac{1}{6}}{\frac{5}{6}} = 0.2\boxed{0.2}

(b)\textbf{(b)}

As we can see, G3G_{3} is the sample space in this case.

G3G_{3} = {4, 5, 6}

Hence, P(G3G_{3}) = P(R4R_{4}) + P(R5R_{5}) + P(R6R_{6})

Hence, P(G3G_{3}) = 16+16+16=36\dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{3}{6}

and R6R_{6} = {6}

Hence, G3R6G_{3} \cap R_{6} = {6} = R6R_{6}

Hence, P(G3R6G_{3} \cap R_{6}) = P(R6R_{6}) = 13\dfrac{1}{3}

Hence, the conditional probability that 6 is rolled given that the roll is greater than 3 i.e.

P(R6G3R_{6} | G_{3}) = P(G3R6)P(G3)\dfrac{P(G_{3} \cap R_{6})}{P(G_{3})} = 1636\dfrac{\frac{1}{6}}{\frac{3}{6}} = 0.33\boxed{0.33}

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