Question 3 of 12

Question 3

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A fair coin tossed 7 times (part a); a bag of 19 marbles: 9 red, 6 blue, 4 white, 3 drawn without replacement (part b).

Find. (a1) $P(\text{4 heads}\mid\text{toss 1 = H})$. (a2) $P(\text{4 heads}\mid\text{toss 1 = H and toss 7 = H})$. (b) $P(\text{more red than white among the 3 drawn})$.

Approach. Conditioning on known tosses shrinks the sample space to the remaining unconstrained tosses; count binomial outcomes there. For the marbles, enumerate all $(r,w)$ count pairs with $r>w$ and sum hypergeometric terms.

1. a1) Condition on toss 1 = H. With toss 1 fixed as heads, 4 heads total requires exactly 3 heads among the remaining 6 (unconstrained, fair) tosses: $$P = \frac{\binom{6}{3}}{2^{6}} = \frac{20}{64}=\frac{5}{16}$$ $\boxed{P=5/16=0.3125}$
2. a2) Condition on toss 1 = H and toss 7 = H. Two of the four required heads are already fixed, so the remaining 5 tosses (2 through 6) need exactly 2 heads: $$P = \frac{\binom{5}{2}}{2^{5}} = \frac{10}{32}=\frac{5}{16}$$ $\boxed{P=5/16=0.3125}$ — the same value as (a1), since in both cases the count reduces to "choose the free-toss positions of the remaining needed heads out of the remaining free tosses," and $\binom{6}{3}/2^6=\binom{5}{2}/2^5$ numerically here.
3. b) Enumerate draws with more red than white. Draw 3 marbles from 19 (9R, 6B, 4W) without replacement; total ways $\binom{19}{3}=969$. Sum over $(r,w)$ with $r>w$, $r+w\le 3$, $b=3-r-w\ge 0$: $$\text{favourable}=\sum_{r>w}\binom{9}{r}\binom{6}{3-r-w}\binom{4}{w}$$ The qualifying triples are $(r,w,b)\in\{(1,0,2),(2,0,1),(3,0,0),(2,1,0)\}$: $$\binom{9}{1}\binom{6}{2}=135,\quad \binom{9}{2}\binom{6}{1}=216,\quad\binom{9}{3}=84,\quad\binom{9}{2}\binom{4}{1}=144$$ Sum: $135+216+84+144=579$. $$P=\frac{579}{969}=\frac{193}{323}$$ $\boxed{P=193/323\approx 0.5975}$