Set Based probability Questions in English

(A) The sample space $S$ consists of $6 \times 6 = 36$ outcomes.
$E_1 = \{(1, 1)\} \implies p_1 = \frac{1}{36}$.
$E_4 = \{(1, 4), (2, 2), (4, 1)\} \implies p_4 = \frac{3}{36} = \frac{1}{12}$.
$E_{10} = \{(2, 5), (5, 2)\} \implies p_{10} = \frac{2}{36} = \frac{1}{18}$.
Comparing the values: $p_1 = \frac{1}{36} \approx 0.0277$,$p_{10} = \frac{2}{36} \approx 0.0555$,$p_4 = \frac{3}{36} \approx 0.0833$.
Thus,$p_1 < p_{10} < p_4$ is correct.

(B) In a non-leap year,the total number of days is $365$.
There are $52$ weeks and $1$ extra day in a non-leap year ($52 \times 7 = 364$ days).
Thus,a non-leap year always has $52$ Sundays.
The remaining $1$ day can be any of the following: Sunday,Monday,Tuesday,Wednesday,Thursday,Friday,or Saturday.
Out of these $7$ possible outcomes,only $1$ outcome is a Sunday.
$\therefore$ Total number of outcomes $= 7$.
Number of favourable outcomes $= 1$.
Hence,the required probability $= \frac{1}{7}$.

(B) The given equation is $ax^{2} + bx + 1 = 0$.
For the equation to have real roots,the discriminant $D$ must be greater than or equal to $0$.
$D = b^{2} - 4ac \geq 0$.
Since $c = 1$,we have $b^{2} - 4a \geq 0$.
The possible values for $(a, b)$ are $(1, 1), (1, 2), (2, 1), (2, 2)$,each with a probability of $1/4$.
Checking the condition $b^{2} - 4a \geq 0$ for each pair:
$1$. For $(1, 1): 1^{2} - 4(1) = -3 < 0$ (Not real).
$2$. For $(1, 2): 2^{2} - 4(1) = 0 \geq 0$ (Real).
$3$. For $(2, 1): 1^{2} - 4(2) = -7 < 0$ (Not real).
$4$. For $(2, 2): 2^{2} - 4(2) = -4 < 0$ (Not real).
Only the pair $(1, 2)$ satisfies the condition.
Thus,the required probability is $1/4$.

(A) Let $X$ be the event that the student is successful. Let $X_1, X_2, X_3$ be the events of passing tests $I, II, III$ respectively.
Given $P(X_1) = p$,$P(X_2) = q$,and $P(X_3) = 1/2$.
The student is successful if $(X_1 \cap X_2)$ or $(X_1 \cap X_3)$.
Since these events are not mutually exclusive,we use the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Here,$A = X_1 \cap X_2$ and $B = X_1 \cap X_3$.
$P(X) = P(X_1 \cap X_2) + P(X_1 \cap X_3) - P(X_1 \cap X_2 \cap X_3)$.
Since the tests are independent:
$P(X) = p \cdot q + p \cdot (1/2) - p \cdot q \cdot (1/2)$.
Given $P(X) = 1/2$:
$1/2 = pq + p/2 - pq/2$.
$1/2 = p/2 + pq/2$.
Multiply by $2$:
$1 = p + pq$.
$1 = p(1+q)$.

(D) Let $A$ be the event of getting an even number on the first die. The possible outcomes for $A$ are $(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)$. Thus,$|A| = 18$.
Let $B$ be the event of getting a sum of $8$. The possible outcomes for $B$ are $(2,6), (3,5), (4,4), (5,3), (6,2)$. Thus,$|B| = 5$.
The intersection $A \cap B$ contains outcomes where the first die is even and the sum is $8$,which are $(2,6), (4,4), (6,2)$. Thus,$|A \cap B| = 3$.
Using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$,we get:
$P(A \cup B) = \frac{18}{36} + \frac{5}{36} - \frac{3}{36} = \frac{20}{36} = \frac{5}{9}$.

(A) Given,$P(A \cap B) = \frac{1}{6}$,$P(A \cup B) = \frac{31}{45}$,and $P(\bar{B}) = \frac{7}{10}$.
Since $P(B) = 1 - P(\bar{B})$,we have $P(B) = 1 - \frac{7}{10} = \frac{3}{10}$.
Using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$,we substitute the values:
$\frac{31}{45} = P(A) + \frac{3}{10} - \frac{1}{6}$.
$P(A) = \frac{31}{45} + \frac{1}{6} - \frac{3}{10} = \frac{62 + 15 - 27}{90} = \frac{50}{90} = \frac{5}{9}$.
Now,check for independence: $P(A) \times P(B) = \frac{5}{9} \times \frac{3}{10} = \frac{15}{90} = \frac{1}{6}$.
Since $P(A \cap B) = P(A) \times P(B)$,the events $A$ and $B$ are independent.

(A) Given that $A$ and $B$ are independent events,$P(A \cap B) = P(A)P(B) = \frac{1}{12} \dots (1)$
The probability that neither $A$ nor $B$ happens is $P(A' \cap B') = \frac{1}{2}$.
By De Morgan's Law,$P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B) = \frac{1}{2}$,so $P(A \cup B) = \frac{1}{2}$.
Using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$,we get:
$P(A) + P(B) - \frac{1}{12} = \frac{1}{2} \implies P(A) + P(B) = \frac{1}{2} + \frac{1}{12} = \frac{7}{12} \dots (2)$
From $(1)$ and $(2)$,$P(A)$ and $P(B)$ are roots of the quadratic equation $x^2 - (P(A)+P(B))x + P(A)P(B) = 0$.
$x^2 - \frac{7}{12}x + \frac{1}{12} = 0 \implies 12x^2 - 7x + 1 = 0$.
Factoring the quadratic: $12x^2 - 4x - 3x + 1 = 0 \implies 4x(3x - 1) - 1(3x - 1) = 0 \implies (4x - 1)(3x - 1) = 0$.
Thus,$x = \frac{1}{4}$ or $x = \frac{1}{3}$.
Therefore,the probabilities are $P(A) = \frac{1}{3}$ and $P(B) = \frac{1}{4}$ (or vice versa).

(C) Let $P(A) = \frac{1}{2}, P(B) = \frac{1}{3}, P(C) = \frac{1}{4},$ and $P(D) = \frac{1}{5}$ be the probabilities of the $4$ students solving the problem individually.
The probability that the problem is solved by at least one student is given by $1 - P(\text{none of the students solve the problem}).$
The probability that a student fails to solve the problem is $P(\bar{X}) = 1 - P(X).$
Thus,$P(\bar{A}) = 1 - \frac{1}{2} = \frac{1}{2}, P(\bar{B}) = 1 - \frac{1}{3} = \frac{2}{3}, P(\bar{C}) = 1 - \frac{1}{4} = \frac{3}{4},$ and $P(\bar{D}) = 1 - \frac{1}{5} = \frac{4}{5}.$
Since the events are independent,the probability that none of them solve the problem is $P(\bar{A} \cap \bar{B} \cap \bar{C} \cap \bar{D}) = P(\bar{A}) \times P(\bar{B}) \times P(\bar{C}) \times P(\bar{D}).$
$P(\text{none}) = \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} = \frac{1}{5}.$
Therefore,the probability that the problem is solved by at least one student is $1 - \frac{1}{5} = \frac{4}{5}.$

(B) The tossing of a coin is an independent event.
Each toss of a fair coin has two equally likely outcomes: Head $(H)$ and Tail $(T)$.
The probability of getting a head in any single toss is $P(H) = \frac{1}{2}$.
The outcome of the fourth toss does not depend on the outcomes of the previous three tosses.
Therefore,the probability of getting a head on the fourth toss remains $\frac{1}{2}$.

(C) Let $S = \{a, b, c, d, e\}$ be a set with $n = 5$ elements.
For any element $x \in S$,there are $4$ possibilities for its membership in the sets $A$ and $B$ such that $A \cap B = \varnothing$:
$1$. $x \in A$ and $x \notin B$
$2$. $x \notin A$ and $x \in B$
$3$. $x \notin A$ and $x \notin B$
(Note: $x \in A$ and $x \in B$ is not allowed as $A \cap B = \varnothing$)
Since there are $5$ elements,the total number of ordered pairs $(A, B)$ is $(2^n) \times (2^n) = 2^5 \times 2^5 = 2^{10} = 4^5$.
The number of favourable pairs $(A, B)$ such that $A \cap B = \varnothing$ is $3^5$ (since each element has $3$ choices).
Thus,the probability $P(E) = \frac{3^5}{4^5} = \frac{3^5}{(2^2)^5} = \frac{3^5}{2^{10}}$.
Comparing this with $\frac{3^p}{2^q}$,we get $p = 5$ and $q = 10$.
Therefore,$p + q = 5 + 10 = 15$.