If you have 2 dice. what is the probability of rolling a 4 at least once after 7 rolls?
+354 The probability of throwing at least one 4 on a pair of dice is 1 in 6, so the probability of sucess is 1/6. The probability of faliure is 1-1/6 which is 5/6. As it is after 7 rolls, there are 7 trials in the experiment.
Putting these values into an equation for binomial distribution, nCr*P^(r)*Q^(n-r), where P is prob of sucess, q is prob of failure, r is number of sucesses and n is the number of trials, we get...
$${\left({\frac{{\mathtt{7}}{!}}{{\mathtt{1}}{!}{\mathtt{\,\times\,}}({\mathtt{7}}{\mathtt{\,-\,}}{\mathtt{1}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{1}}}{{\mathtt{6}}}}\right)}^{{\mathtt{1}}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{5}}}{{\mathtt{6}}}}\right)}^{\left({\mathtt{7}}{\mathtt{\,-\,}}{\mathtt{1}}\right)} = {\mathtt{0.390\: \!714\: \!306\: \!127\: \!114\: \!8}}$$
as the probability of getting a 4 in one roll.
This is very likely wrong.
Anyone better at maths should check this first, since I'm very very unconfident in my binomial skills.
+354 The probability of throwing at least one 4 on a pair of dice is 1 in 6, so the probability of sucess is 1/6. The probability of faliure is 1-1/6 which is 5/6. As it is after 7 rolls, there are 7 trials in the experiment.
Putting these values into an equation for binomial distribution, nCr*P^(r)*Q^(n-r), where P is prob of sucess, q is prob of failure, r is number of sucesses and n is the number of trials, we get...
$${\left({\frac{{\mathtt{7}}{!}}{{\mathtt{1}}{!}{\mathtt{\,\times\,}}({\mathtt{7}}{\mathtt{\,-\,}}{\mathtt{1}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{1}}}{{\mathtt{6}}}}\right)}^{{\mathtt{1}}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{5}}}{{\mathtt{6}}}}\right)}^{\left({\mathtt{7}}{\mathtt{\,-\,}}{\mathtt{1}}\right)} = {\mathtt{0.390\: \!714\: \!306\: \!127\: \!114\: \!8}}$$
as the probability of getting a 4 in one roll.
This is very likely wrong.
Anyone better at maths should check this first, since I'm very very unconfident in my binomial skills.
"2 dice" doesn't make sense unless we're adding them, that is, a 1 and a 3, or two 2s. The probability of that is 3 out of 36 (1-3, 2-2, 3-1) which equals 1 out of 12. The probability of NOT totalling 4 is 11 out of 12. So the probability of not totalling 4, 7 times is (11/12)^7. The probability of at least one 4 is the opposite, 1 - (11/12)^7 which is 0.4561488217396119.
