A die is thrown twice. What is the probability that $(i)$ 3 will not come up either time? $(ii)$ 6 will come up at least once?
- $(i) \displaystyle\frac{12}{25} \\ (ii) \displaystyle\frac{16}{25}$
- $(i) \displaystyle\frac{18}{25} \\ (ii)\displaystyle\frac{9}{25}$
- $(i) \displaystyle\frac{25}{36} \\ (ii) \displaystyle\frac{11}{36}$
- $(i) \displaystyle\frac{23}{36} \\ (ii) \displaystyle\frac{17}{36}$
(i) Probability of 3 not coming up in one throw is 5/6. For two throws, (5/6)^2 = 25/36. (ii) Probability of 6 coming up at least once = 1 - P(no 6) = 1 - (5/6)^2 = 1 - 25/36 = 11/36.
For the first part, the probability of not rolling a 3 on a single die is 5 divided by 6, so the probability of not getting a 3 on either of two rolls is 5/6 multiplied by 5/6, which equals 25/36. For the second part, it is easier to use the complement method; the probability of not rolling a 6 on either roll is 5/6 multiplied by 5/6, equaling 25/36. The probability of getting at least one 6 is 1 minus 25/36, which is 11/36. The probabilities are 25/36 and 11/36, giving the result (i) 25/36 and (ii) 11/36.