Probability

\(2/9\)

(i) The probability that none of the first three rolls have a 4 is \( (5/6)^3 \). So the required probability is \(1 - (5/6)^3\).

(ii) Let \(S\) be the number of ways in which four rolls can have exactly two multiples of 3. The required probability \(P\) is then \(S/6^4\).

Let us calculate \(S\). Two positions can be picked in \( \binom{4}{2} \) ways. These two positions can have either a 3 or a 6. So the favorable positions can be filled in \( \binom{4}{2}\times 2^2 \) ways. The other two positions can have either 1, 2, 4 or 6. So they can be filled in \( 4^2\) ways. Hence we have: \[ P = \frac{ \binom{4}{2}\times 4 \times 4 \times 4 }{6^4} = \frac{2}{3}^3 = \frac{8}{27} \]

1. Since the outcome of the toss was tails, the second daugther lost since the die came up 1,2,3 or 4. Hence, required probability is 2/3.
2. Let \(A\) the event that the second daughter won and \(B\) be the event that the first daughter won.
   \[ P(A \mid B)=\frac{P(A \cap B)}{P(B)} \] \begin{align*} P(A \cap B)&=P(A \cap B \mid H) P(H)+P(A \cap B \mid T) P(T) \\ &=1\cdot \frac{1}{2} + \frac{1}{9}\cdot \frac{1}{2} \\ &=\frac{5}{9} \end{align*} \begin{align*} P(B)&=P(B \mid H) P(H)+P(B \mid T) P(T) \\ &=1\cdot \frac{1}{2} + \frac{1}{3}\cdot \frac{1}{2} \\ &=\frac{2}{3} \end{align*} \[ P(A \mid B)=\frac{P(A \cap B)}{P(B)} = \frac{5/9}{2/3} = \frac{5}{6} \]

(a) The probability of Tinku not getting a chocolate in one step is \(1-\frac{1}{k}\).

\begin{align} \mathrm{P}(\text { Tinku gets at least one chocolate })&=1-\mathrm{P}(\text { Tinku gets none }) \\ &=1-\left(1-\frac{1}{k}\right)^{n} \end{align}

Sol 1. (b) There are \(\left(\begin{array}{c}(n-1)+k-1 \\ k-1\end{array}\right)\) distributions in which Tinku gets at least a chocolate: give Tinku a chocolate and then use the given formula to find number of distributions of the remaining \(n-1\) chocolates among \(k\) students. So the answer is \(\left(\begin{array}{c}(n-1)+k-1 \\ k-1\end{array}\right) /\left(\begin{array}{c}n+k-1 \\ k-1\end{array}\right)=\frac{n}{n+k-1} \)

Sol 2. (b) The number of distributions in which Tinku gets no chocolate \(=\) number of distributions of \(n\) chocolates among the remaining \(k-1\) students \(=\left(\begin{array}{c}n+k-2 \\ k-2\end{array}\right) .\) So the desired probability is \(1-\left(\begin{array}{c}n+k-2 \\ k-2\end{array}\right) /\left(\begin{array}{c}n+k-1 \\ k-1\end{array}\right)=\frac{n}{n+k-1}\)

(a) The equation \(x^{2}+b x+c=0\) has a real solution if and only if \(b^{2}-4 c\) is \(\geq 0\)

(b) The value of \(p\left(\frac{1}{2}\right),\) i.e., the probability that \(x^{2}+\frac{x}{2}+c=0\) has a real solution is \(\frac{1}{16}\). There is a real root when \(b^{2}-4 c=\frac{1}{4}-4 c \geq 0,\) i.e., \(0 \leq c \leq \frac{1}{16}\) which is \(\frac{1}{16}\) fraction of the interval [0,1] over which \(c\) ranges.

(c) Increasing. This is because \(b^{2}-4 c \geq 0\) if and only if \(0 \leq c \leq \frac{b^{2}}{4}\).
So \(p(b)=\frac{b^{2}}{4},\) which is increasing for \(0 \leq b \leq 1\)

(d) This is the fraction of the area of the unit square \([0,1] \times[0,1]\) that is occupied by the region \(b^{2}-4 c \geq 0,\) i.e., it is the area under the parabola \(c=\frac{b^{2}}{4}\) from \(b=0\) to \(b=1\) which is \(\int_{0}^{1} \frac{b^{2}}{4} d b=\frac{1}{12}\)

(i) \(\operatorname{Pr}\left[E_{4}\right]=\frac{4 !}{4^{4}}=\frac{3}{32}\) (ii) \(\operatorname{Pr}\left[E_{4} \mid E_{3}\right]=\frac{1}{4}\) (iii) \(\operatorname{Pr}\left[E_{4} \mid E_{2}\right]=\frac{24}{4^{2}}=\frac{1}{8}\) (iv) \(\operatorname{Pr}\left[E_{3} \mid E_{4}\right]=1\)

\(1/33\)

(a) The probability that a randomly chosen student is well prepared is \(5 / 14\) The probability of a well prepared student answering two randomly chosen questions correctly is \(\left(\begin{array}{l}9 \\ 2\end{array}\right) /\left(\begin{array}{l}10 \\ 2\end{array}\right) .\) So the probability that a randomly chosen student is well prepared AND answers two randomly chosen questions correctly is \(\frac{5}{14} \times \frac{\left(\frac{9}{20}\right)}{(10)}=\frac{2}{7} .\) A student belongs to exactly one of the three preparedness categories, so the desired probability is obtained by adding \(\frac{2}{7}\) with the results of parallel calculations for the other two categories. We get

Let \(P(W),P(M)\) and \(P(K)\) denote the probability that the student is well prepared, moderately prepared and poorly prepared, respectively.

\[ P(\text { both answers correct })= P(\text { W }) \frac{\left(\begin{array}{l} 9 \\ 2 \end{array}\right)}{\left(\begin{array}{c} 10 \\ 2 \end{array}\right)}+P(\text { M }) \frac{\left(\begin{array}{c} 6 \\ 2 \end{array}\right)}{\left(\begin{array}{c} 10 \\ 2 \end{array}\right)}+P(\text { K }) \frac{\left(\begin{array}{c} 3 \\ 2 \end{array}\right)}{\left(\begin{array}{c} 10 \\ 2 \end{array}\right)} \] which equals \[ \frac{5}{14} \times \frac{36}{45}+\frac{6}{14} \times \frac{15}{45}+\frac{3}{14} \times \frac{3}{45}=\frac{31}{70} \]

(b) The probability that a randomly chosen student was well prepared given that he answered both questions correctly is

\begin{align} P(\text { well prepared | both correct })&=\frac{P(\text { well prepared and both correct })}{P(\text { both correct })}\\ &=\frac{2 / 7}{31 / 70}=\frac{20}{31} \end{align}