Hi @yi_fan_bao,
First, welcome to our community!
It would be \frac{1}{2} if we ask for the probability of getting a 6 if we know that the dice result is greater or equal 4. However, the experiment is a bit different: You throw one dice, if the result is greater or equal to 4, then you throw another dice , and use this result. The second dice can have any value between 1 and 6.
I will rewrite the question so it become a bit more clear about the experiment.
Was this clear? Please let me know if you still have questions about it.
Thanks,
Lucas
Hi Lucas,
I’m still unclear about how to calculate the probabilities.
P(\text{getting a }6| E < 4) = \frac{\text{No. of possibilities of getting a 6 given } E< 4} {\text{Total No. of all possible values given } E<4}
Given that 1st coin is restricted to values 1, 2, 3. The possible values of 2nd coin that would make the sum to 6 are 5, 4, 3. These three are the possibilities (1,5) (2,4) (3,3).
Total number of possible values given E<4. For first die it’s (1,2,3) 3 values & second die (1,2,3,4,5,6) 6 values. Total 3*6 = 18 possibilities.
P(\text{getting a 6}| E < 4) = \frac{3}{18} = \frac{1}{6}
P(\text{getting a 6}| E \ge 4) = \frac{\text{No. of possibilities of getting a 6 given }E \ge 4} {\text{Total No. of all possible values given }E\ge4}
This is the case of single die toss. So
P(\text{getting a } 6| E \ge 4) = \frac{1}{6}
So,
P(\text{getting a }6) = P(\text{getting a } 6| E < 4) + P(\text{getting a } 6| E \ge 4)
P(\text{getting a }6) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}
But there is no option \frac{1}{3} in the list of options. Please help me understand how to calculate these probabilities.
@lucas.coutinho Please respond.
Sorry for the late response.
I see your point. Your calculations would be completely correct if we considered the value for the first dice! However, the game is as follows:
You throw 1 dice. If the result is less than 4, then you throw 1 dice, otherwise, you throw 2 dice. The first throwing of a dice in the game is just to decide whether you will throw one or two dice in the game.
Regards,
Lucas
Hi @lucas.coutinho,
Thanks for your response. And sorry for the late reply. It all makes sense after reading your explanation. I’ve identified the way to arrive at the right answer 11/72. 