Question

Ques. A biased die has a probability of 1/4 of showing a 5, while the probability of any of 1, 2, 3, 4, or 6 turning up is the same . If three such dice are rolled, what is the probability of getting a sum of atleast 14 without getting a 6 on any die ?

Answer 1

1/64 + 3/320 + 3/320 + 3/320 = 7/160

Answer: 7/160

Answer 2

Since the addition of every probable event is equal to 1 and the probability of rolling a five is 1/4, then the probability of any of the other numbers is:

$\text{P(not five): }1 - \frac{1}{4} = \frac{3}{4}$
$\text{P(1) = P(2) = P(3) = P(4) = P(6): } \frac{3}{20}$

Now, the condition is that no die rolls a six. Thus, we need to limit our sample space from 216 down to 125, to ensure we're not rolling a six.

The ways we can reach 14 or greater are:
5, 5, 5
4, 5, 5 (which can occur in 3 ways)

Thus, our probability becomes:
$(\frac{1}{4})^{3} + 3 \cdot \frac{3}{20} \cdot (\frac{1}{4})^{2}$
$= \frac{1}{64} + \frac{9}{320}$

$= \frac{7}{160}$