# Die and toss random variable probability question

Here is the question:

There are two die. Blue and Green. Fair blue die has sides 1,1,3,3,5,5. B is a random variable which represents the score when blue die is rolled. G represents score shown on green die. And P(G=g) is :

g = 2 ; P(g)= 2/3

g = 4 ; P(g)= 1/6

g = 6 ; P(g)= 1/6


Student A1 flips a fair coin that has sides 2 and 5. Student A2, seeing number on coin chooses a die to roll. Student A2 must score greater than flip on coin to win. Otherwise Student A1 wins. Find the probability that Student A2 wins.

My approach to solve this question is :

I have to find Probability mass function for B 1/3 for each 1,3,5. {Since there are two for each number so 2/6 for each} Now, Probability for coin is:

P(2)=P(5)= 0.5 {Since coin is fair}

If coin flip results in 2 then Student A2 will choose Blue die because blue has a higher chance of getting a value greater than 2. And if coin flip results in 5 then Student A2 must choose Green die. $$P(Student A2 wins) ={ (P(Coin =2 \hat B=3) | P(Coin =2 \hat B=5)) /P(B)} +( P(Coin =5 \hat G=6)) /P(G)$$

Is this correct?

## 2 Answers

The probability of student A2 winning is:

(probability that $$coin = 2$$)(probability of $$B > 2$$) + (probability that $$coin = 5$$)(probability of $$G > 5$$) $$= 0.5*(2/3) + 0.5*(1/6) = 1/3 + 1/12 = 5/12$$

Your logic seems correct until the last statement (the equation of probability). There would not be a division of probability of coin by the probability of either die. Since student A2's consideration of B or G is conditional to the coin outcome, the probability would be multiplicative.

• Why is it being multiplied by 2/3. Shouldn't it be 0.3^2 – puffles Oct 2 '18 at 6:37
• Blue die had 4 sides that give greater than 2 out of 6 sides. Thus the probability is 4/6 which is 2/3 – Kenny Kim Oct 2 '18 at 13:29

$$P(\text{A2 wins})=P(\text{A2 wins}\mid C=2)P(C=2)+P(\text{A2 wins}\mid C=5)P(C=5)=$$$$P(B>2)P(C=2)+P(G>5)P(C=5)=\frac46\frac12+\frac16\frac12=\frac5{12}$$