(a) A professor designed his final exam as follows: There will be three sections in the exam. Each section has five questions. Students have to pick any two sections to answer, in any order. Within each section, they must choose any three questions. In how many possible ways can a student choose which questions to answer?

(b) A committee of five people is to be chosen from a club that has ten scientists and eight engineers. How many ways can the committee be formed if it has to contain at least two scientists and at least one engineer?

(c) Suppose you are choosing ve cards at random from a standard deck. What is the probability that there will be the ace of spades, and at least one card each of the three other suits? Please explain the reasoning behind your calculations.

(d) Suppose ten coins are being flipped one after another, and the outcome of each toss noted down. What is the probability that every alternate toss will be a head? Please explain the reasoning behind your calculations.

(a) $\binom{3}{2} \binom{5}{3}$

(b) $\binom{10}{2} \binom{8}{3} + \binom{10}{3}\binom{8}{2} + \binom{10}{4}\binom{8}{1}$

(c) $\dfrac{\binom{13}{1} \binom{13}{1} \binom{13}{2} * 3}{\binom{52}{5}}$

One from clubs, one from hearts and one from diamonds - $\binom{13}{1} \binom{13}{1} \binom{13}{2}$
One from diamonds, one from hearts and one from clubs - $\binom{13}{1} \binom{13}{1} \binom{13}{2}$
One from clubs, one from diamonds and one from hearts - $\binom{13}{1} \binom{13}{1} \binom{13}{2}$
By the sum rule, multiply by 3.
Total number of possibilities = $\binom{52}{5}$

(d) $\dfrac{2}{2^{10}}$

Only two events possible: HTHTHTHTHT or THTHTHTHTH among the $2^{10}$ possibilities.

Do you think my answers are correct?

  • $\begingroup$ The first is probably right idea but wrong answer. Want $\binom{3}{2}\binom{5}{3}\binom{5}{3}$. Note that writing down an expression is not an explanation. $\endgroup$ – André Nicolas Apr 21 '14 at 20:43
  • $\begingroup$ (b) and (c) look correct, but for (d): Specifying that all alternate tosses be heads does not require that all the other tosses be tails. $\endgroup$ – Graham Kemp Apr 21 '14 at 21:26
  • $\begingroup$ Correction: (c) is off, as noted below we neglected the case of one from each other suit and another spade. $\endgroup$ – Graham Kemp Apr 21 '14 at 23:11

As noted above, (a) is incorrect, it should be ${3\choose 2}{5\choose 3}{5 \choose 3}$.

(b) is correct. Putting it in that form shows your thinking, but more explicit explanation might be desired.

(c) is wrong. You are not taking into account the possibility that the fifth card is a spade. Also, your wording is off, I think you mean: "One from clubs, one from hearts, and two from diamonds", etc.

(d), what exactly is meant by "every alternate toss will be a head"? Does that mean heads and tails alternate or that all the odd-number (or all the even-number) tosses are heads? I favor the second meaning for that particular word choice, which gives a different answer. I would give both answers and explain which goes with which interpretation of the question.


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