Probability

Use Bayes’ theorem, with \(F\) = the fake coin is chosen and \(HH\) = two heads.

\(P(F) = \frac{1}{5}\) and \(P(HH \mid F) = 1\); a regular coin is chosen with probability \(\frac{4}{5}\) and gives two heads with probability \(0.5 \times 0.5 = \frac{1}{4}\).

\(P(HH) = \frac{1}{5} \cdot 1 + \frac{4}{5} \cdot \frac{1}{4} = \frac{2}{5}\), so \(P(F \mid HH) = \dfrac{1/5}{2/5} = \frac{1}{2} = 0.50\).

Answer: 0.50

The total probability must be 1, which fixes \(C\):

\[\int_1^4 Cx^2\,dx = C \cdot \frac{4^3 - 1^3}{3} = 21C = 1 \;\Rightarrow\; C = \frac{1}{21}\]

\[P(2 \le x \le 3) = \frac{1}{21}\int_2^3 x^2\,dx = \frac{1}{21} \cdot \frac{3^3 - 2^3}{3} = \frac{19}{63} \approx 0.3016\]

Rounded to three decimal places, this is 0.302.

Answer: 0.302

Square invariance means the roots \(\{\alpha, \beta\}\) and \(\{\alpha^{2}, \beta^{2}\}\) are the same pair. There are two ways to match them:

• \(\alpha^{2} = \alpha\) and \(\beta^{2} = \beta\): each root is 0 or 1, giving \(x^{2}\), \(x(x - 1)\) and \((x - 1)^{2}\).
• \(\alpha^{2} = \beta\) and \(\beta^{2} = \alpha\): then \(\alpha^{4} = \alpha\), so \(\alpha \in \{0, 1, \omega, \omega^{2}\}\), where \(\omega\) is a primitive cube root of unity. The only new polynomial is \((x - \omega)(x - \omega^{2}) = x^{2} + x + 1\).

So there are exactly 4 square invariant quadratics, and 2 of them, \(x^{2}\) and \((x - 1)^{2}\), have equal roots. The probability is \(2/4 = 0.5\).

Answer: 0.5

Let the cut be at \(U\), uniform on \((0,1)\). If \(U > 0.4\), the piece containing 0.4 is \((0, U)\), of length \(U\); if \(U < 0.4\), it is \((U, 1)\), of length \(1 - U\).

\[E = \int_{0}^{0.4} (1 - u)\,du + \int_{0.4}^{1} u\,du = (0.4 - 0.08) + \frac{1 - 0.16}{2} = 0.32 + 0.42 = 0.74\]

The result is above 0.5 because a fixed point is more likely to land in the longer of the two pieces.

Answer: 0.74

By symmetry, 1 comes before 2 in exactly half of all permutations, so \(P(X) = \frac{1}{2}\); likewise \(P(Y) = \frac{1}{2}\).

The order inside the pair \(\{1, 2\}\) and the order inside the pair \(\{3, 4\}\) are each equally likely either way and do not influence each other, so \(P(X \cap Y) = \frac{1}{4} = P(X)\,P(Y)\). Hence \(X\) and \(Y\) are independent.

• A fails: both events can happen together (probability \(\frac{1}{4}\)).
• C fails: both can fail together, for example in 2, 1, 4, 3, ….
• D fails: the two events are equally likely.

Answer: B

• A: false. Independence needs \(P(A \cap B) = P(A)P(B) = 0.3 \times 0.5 = 0.15\), but \(P(A \cap B) = 0.1\).
• B: true. \(P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.3 + 0.5 - 0.1 = 0.7\).
• C: true. \(A\) splits into the disjoint parts \(A \cap B\) and \(A \cap B^{c}\), so \(P(A \cap B^{c}) = 0.3 - 0.1 = 0.2\).
• D: false. By De Morgan's law \(A^{c} \cap B^{c} = (A \cup B)^{c}\), so \(P(A^{c} \cap B^{c}) = 1 - 0.7 = 0.3\), not 0.4.

Answer: B, C

Let \(R_{1}\) and \(R_{2}\) be the events that the first and the second ball are red. Then

\[P(R_{1} \cap R_{2} \mid R_{1}) = \frac{P(R_{1} \cap R_{2})}{P(R_{1})} = \frac{\frac{10}{25} \times \frac{9}{24}}{\frac{10}{25}} = \frac{9}{24} = 0.375\]

In words: once a red ball is out, the bag holds 9 red and 15 blue balls, so the second ball is red with probability \(9/24\). The unconditional probability of two reds, \(\frac{10}{25} \times \frac{9}{24} = 0.15\), answers a different question.

Answer: 0.375

Each die shows one of 6 faces independently, so there are \(6^{6} = 46656\) equally likely outcomes.

For all six numbers to be distinct, the dice must show 1 to 6 in some order. The first die can take any of 6 values, the second any of the remaining 5, and so on, giving \(6! = 720\) favourable outcomes.

\[P = \frac{6!}{6^{6}} = \frac{720}{46656} = \frac{5}{324}\]

(Dividing 720 and 46656 by their common factor 144 gives 5 and 324.)

Answer: B

Because \(0 \le y \le 1\) and \(x \ge 0\), every outcome has \(z = xy \le x\). Taking expectations keeps the inequality, so \(\overline{z} \le \overline{x}\) always holds.

Options (A) to (C) compare \(E[xy]\) with \(E[x]\,E[y]\); the difference is the covariance of \(x\) and \(y\), which can have either sign when they are dependent. Take \(x\) uniform on \([0,1]\):

• \(y = x\): \(\overline{z} = E[x^{2}] = \tfrac{1}{3} > \tfrac{1}{4} = \overline{x}\,\overline{y}\), so (A) and (B) fail.
• \(y = 1 - x\): \(\overline{z} = \tfrac{1}{2} - \tfrac{1}{3} = \tfrac{1}{6} < \tfrac{1}{4} = \overline{x}\,\overline{y}\), so (C) fails.

Answer: D