Algebra

Probability

Medium

applied

The probability that in a year of the 22nd century choosen at random, there will be 53 Sundays is:

$\frac{3}{28}$

$\frac{2}{7}$

$\frac{5}{28}$

$\frac{1}{7}$

✅ Correct Option: 3

Every year guarantees 52 complete weeks (= 52 Sundays). The only question is whether the extra leftover days include a Sunday.

An ordinary year has 365 days = 52 weeks + 1 extra day.

A leap year has 366 days = 52 weeks + 2 extra days.

Since every year already has 52 Sundays, 53 Sundays occur only if at least one of the extra days is a Sunday.

In any century of 100 years:

$\text{Leap years} = \dfrac{100}{4} = 25$

$\text{Ordinary years} = 100 - 25 = 75$

An ordinary year has 1 extra day, which can be any one of: {Sun, Mon, Tue, Wed, Thu, Fri, Sat}

For 53 Sundays, that 1 extra day must be Sunday.

$P(\text{53 Sundays | Ordinary Year}) = \dfrac{1}{7}$

A leap year has 2 extra days, which are consecutive. The possible pairs are:

(Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun)

Out of 7 possible pairs, 2 pairs contain a Sunday → (Sun, Mon) and (Sat, Sun)

$P(\text{53 Sundays | Leap Year}) = \dfrac{2}{7}$

Using the Law of Total Probability:

$P(\text{53 Sundays}) = P(\text{Ordinary}) \times P(\text{53 Sun | Ordinary}) + P(\text{Leap}) \times P(\text{53 Sun | Leap})$

$= \dfrac{75}{100} \times \dfrac{1}{7} + \dfrac{25}{100} \times \dfrac{2}{7}$

$= \dfrac{75}{700} + \dfrac{50}{700}$

$= \dfrac{125}{700}$

$= \boxed{\dfrac{5}{28}}$

2026: 31 May Shift 2

If $A$ and $B$ are two events such that $P(\overline{A} \cup \overline{B}) = \dfrac{2}{3}$ and $P(A \cup B) = \dfrac{5}{9}$ , then which of the following are TRUE?

A. $P(A \cap B) = \dfrac{1}{3}$

B. $P(A \cup B) + P(A \cap B) = \dfrac{7}{9}$

C. $P(A) + P(B) = \dfrac{2}{9}$

D. $P(\overline{A}) + P(\overline{B}) = \dfrac{10}{9}$

Choose the correct answer from the options given below:

2025: 26 May Shift 2

If the events A and B are independent, then which of the following statements are true?

(A) P(A'B) = [1-P(A)] P(B)

(B) A and B are mutually exclusive

(D) P(A'B') = [1-P(A)] [1-P(B)]

Choose the correct answer from the options given below:

2025: 15 May Shift 1

If a computer code is correctly programmed, it gives 90% acceptable results. But if it is not correctly programmed, it gives only 40% acceptable results. From previous experience, it is observed that only 80% of codes are correctly programmed. If after a certain programming, the code gives 2 acceptable results, then the approximate probability that the code is correctly programmed is

2025: 14 May Shift 2

If a person A speaks the truth in 80% cases and the person B speaks the truth in 75% cases, then the probability that they contradict each other in a statement is

2026: 31 May Shift 2

In a factory of fans, machine A, B and C manufacture respectively 20%, 40% and 40% of total fans. Out of their output 2, 3 and 5 percent are defective fans respectively. A fan is drawn at random from the products then the probability that the drawn fan is defective is:

2022: 17 Aug Shift 2

Match List - I with List - II.

List - I	List - II
(A) Two events E and F will be independent if $P(E'F')$ is equal to	(I) $1 - P(E/F)$
(B) If $P(F) \neq 0$ , then $P(E'/F)$ is equal to	(II) $P(E) = P(F)$
(C) If E and F are independent events, then	(III) $P(E \cap F') = P(E) \cdot P(F')$
(D) If $P(E \cap F) \neq 0$ and $P(E/F) = P(F/E)$ , then	(IV) $[1-P(E)][1-P(F)]$

Choose the correct answer from the options given below :

The probability that in a year of the 22nd century choosen at random, there will be 53 Sundays is:

Solution

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