STAT/MATH 394A

1. (a) There are 9 choices for the hundreds place, then 9 choices for the tens place that are different from the hundreds place, and finally 8 choices for the units that are different from the previous two digits. Hence, by the basic counting rule, there are 9 x 9 x 8 = 648 integers with distinct digits.

(b) There are 5 choices for the units to make the number odd, then 8 choices for the hundreds place that are different from the units, and finally 8 choices for the tens different from the other two digits. Hence 5 x 8 x 8 = 320 odd integers with distinct digits. Note that less than half of the integers with distinct digits are odd.

2.20x10x18x9x16x8x...x2x1/20! =

3. (a) P(A₁)=P(S)=1 so P(A_k)=1/k. Hence P(e_k)=P(A_k)-P(A_k+1)=1/k - 1/(k+1) = 1/k(k+1). This specifies all probabilities.

(b)

4.1/6 x 3/6 (even on roll 2) + 2/6 (even not 6 on roll 1)x 3/6 x 3/6 = 1/6 .

5. A straightforward application of Bayes' theorem yields P(A₁|rain)=.05x.2/(.05 x .2+.4 x .5+.9 x .3)=.02

P(A₂|rain)=.20/.48=.42 and P(A₃|rain)=.27/.48=.56.