For part b) below, is the probability equal to $(\frac{1}{6}+\frac{1}{6}-\frac{1}{36})$, or $(\frac{1}{6}+\frac{1}{6}-\frac{1}{30})$? I think it should be-$\frac{1}{30}$, but the answers online say it should be-$\frac{1}{36}$. Could anyone explain why?
Six runners are entered in a track meet, and have equal ability. what is the probability that
a) they will finish in ascending order of their ages?
b) Shanaze will finish first or Tanya will finish second?
c) Shanaze and Tanya will not finish back-to-back?
b) A-One fixed runner (Shanaze) is first with probability $\frac{1}{6}$ (cause one of six runners must take the first place) Similar,B - (Tanya) will finish second with probability $\frac{1}{6}$ (reasoning as above)
And now, to calculate $P(A \cup B) = P(A) + P(B) - P(A \cap B)$, we need $P(A \cap B)$ That is the probability of Shaneze finishing first and Tanya finishing second simultaneously. Clearly our fixed runners are placed in only 1 way (Sha first, Tan second), then we have 4 runners to order, which gives us 4! ways. Now, all ways of arranging 6 runners is 6!, so our probability is $P(A \cap B) = \frac{|A \cap B|}{|\Omega|} = \frac{4!}{6!} = \frac{1}{5 \cdot 6} = \frac{1}{30} $
Hence, the answer is $P(A\cup B)= \frac{1}{6} + \frac{1}{6} - \frac{1}{30} = \frac{9}{30} = \frac{3}{10} $