I tried to solve the exercise and it seems meaningful, but the problem is that in the book the result is $25/72$, while mine is $1/7776$. I don't know if maybe I didn't understand the problem, but by the text I understood that there are two dices rolled three times, and I have to calculate the probability to have 7 as result. Now, I could use the binomial probability density function, but I didn't know very well how to use the data I have (if someone can also explain me the resolution with this structure I'll thank him). By logic, throwing three times two dices, the total number of possibilities is 46'656, because I have $36^3$. Now, to have 7 as result and a dice must have at least 1 as value, the only possibility that I have to reach 7 is 11/11/12 (the first two dices are 1, the 3th and 4th 1, the 5th 1 and the 7th 2) and all the combination of this sequence: 11/11/12 11/11/21 11/21/11 11/12/11 21/11/11 12/11/11

Calculating the probability, I have $3* (1/36 * 1/36 * 2/36)$, and the probability of having 7 as result is $6/7776$

I don't know where I'm wrong.


1 Answer 1


To get the result you can start by calculating the probability of obtaining a sum of 7 when rolling two dices: P(7)= 6/36 = 1/6

And then as you say its a binomial: So it's combinatory of 1 in 3 times the probability of success to the power of 1 times probability of failure to the power of 2 So: (3!/1!*(3-1)!) * (1/6)^1 * (5/6)^2

Hope it makes everything clear

  • $\begingroup$ But I don't understand why I have to put 3 choose 1, because I know that 3 is the total and 1 is the success $\endgroup$
    – Ele975
    Dec 29, 2020 at 21:46

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