What is the probability that a randomly chosen leap year has 52 Sundays? 1. 2. 3. 4.

Question

KnowledgeBoat · Accepted Answer

In a leap year, there are 366 days.

366 days = 52 weeks + 2 days

The 7 possible pairs for the 2 extra days are:(Monday, Tuesday)(Tuesday, Wednesday)(Wednesday, Thursday)(Thursday, Friday)(Friday, Saturday)(Saturday, Sunday)(Sunday, Monday)

Total number of possible outcomes = 7

Number of outcomes where a Sunday occurs = 2 [(Saturday, Sunday) and (Sunday, Monday)]

Number of outcomes where a Sunday does not occur = 7 - 2 = 5. {i.e,(Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat)}

Let E be the event that a leap year has exactly 52 Sundays,

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{5}{7}$

Hence, option 3 is the correct option.

Mathematics

What is the probability that a randomly chosen leap year has 52 Sundays?
$\Big(\dfrac{1}{7}\Big)$
$\Big(\dfrac{2}{7}\Big)$
$\Big(\dfrac{5}{7}\Big)$
$\Big(\dfrac{6}{7}\Big)$

Probability

Answer

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A card is drawn at random from a pack of 52 playing cards. The probability of drawing a card which is neither a spade nor a king, is:
$\Big(\dfrac{17}{52}\Big)$
$\Big(\dfrac{4}{13}\Big)$
$\Big(\dfrac{35}{52}\Big)$
$\Big(\dfrac{9}{13}\Big)$

A card is drawn at random from a well-shuffled pack of 52 cards. The probability that the drawn card is neither a king nor a queen, is:
$\Big(\dfrac{2}{13}\Big)$
$\Big(\dfrac{3}{13}\Big)$
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The probability that a leap year has 53 Sundays is:
$\Big(\dfrac{1}{7}\Big)$
$\Big(\dfrac{2}{7}\Big)$
$\Big(\dfrac{3}{7}\Big)$
$\Big(\dfrac{4}{7}\Big)$

The probability of a non-leap year having 53 Mondays is:
$\Big(\dfrac{1}{7}\Big)$
$\Big(\dfrac{2}{7}\Big)$
$\Big(\dfrac{5}{7}\Big)$
$\Big(\dfrac{6}{7}\Big)$

Mathematics

What is the probability that a randomly chosen leap year has 52 Sundays?(17)\Big(\dfrac{1}{7}\Big)(71​)(27)\Big(\dfrac{2}{7}\Big)(72​)(57)\Big(\dfrac{5}{7}\Big)(75​)(67)\Big(\dfrac{6}{7}\Big)(76​)

Probability

Answer

Related Questions

A card is drawn at random from a pack of 52 playing cards. The probability of drawing a card which is neither a spade nor a king, is:(1752)\Big(\dfrac{17}{52}\Big)(5217​)(413)\Big(\dfrac{4}{13}\Big)(134​)(3552)\Big(\dfrac{35}{52}\Big)(5235​)(913)\Big(\dfrac{9}{13}\Big)(139​)

A card is drawn at random from a well-shuffled pack of 52 cards. The probability that the drawn card is neither a king nor a queen, is:(213)\Big(\dfrac{2}{13}\Big)(132​)(313)\Big(\dfrac{3}{13}\Big)(133​)(1013)\Big(\dfrac{10}{13}\Big)(1310​)(1113)\Big(\dfrac{11}{13}\Big)(1311​)

The probability that a leap year has 53 Sundays is:(17)\Big(\dfrac{1}{7}\Big)(71​)(27)\Big(\dfrac{2}{7}\Big)(72​)(37)\Big(\dfrac{3}{7}\Big)(73​)(47)\Big(\dfrac{4}{7}\Big)(74​)

The probability of a non-leap year having 53 Mondays is:(17)\Big(\dfrac{1}{7}\Big)(71​)(27)\Big(\dfrac{2}{7}\Big)(72​)(57)\Big(\dfrac{5}{7}\Big)(75​)(67)\Big(\dfrac{6}{7}\Big)(76​)

What is the probability that a randomly chosen leap year has 52 Sundays?
$\Big(\dfrac{1}{7}\Big)$
$\Big(\dfrac{2}{7}\Big)$
$\Big(\dfrac{5}{7}\Big)$
$\Big(\dfrac{6}{7}\Big)$

A card is drawn at random from a pack of 52 playing cards. The probability of drawing a card which is neither a spade nor a king, is:
$\Big(\dfrac{17}{52}\Big)$
$\Big(\dfrac{4}{13}\Big)$
$\Big(\dfrac{35}{52}\Big)$
$\Big(\dfrac{9}{13}\Big)$

A card is drawn at random from a well-shuffled pack of 52 cards. The probability that the drawn card is neither a king nor a queen, is:
$\Big(\dfrac{2}{13}\Big)$
$\Big(\dfrac{3}{13}\Big)$
$\Big(\dfrac{10}{13}\Big)$
$\Big(\dfrac{11}{13}\Big)$

The probability that a leap year has 53 Sundays is:
$\Big(\dfrac{1}{7}\Big)$
$\Big(\dfrac{2}{7}\Big)$
$\Big(\dfrac{3}{7}\Big)$
$\Big(\dfrac{4}{7}\Big)$

The probability of a non-leap year having 53 Mondays is:
$\Big(\dfrac{1}{7}\Big)$
$\Big(\dfrac{2}{7}\Big)$
$\Big(\dfrac{5}{7}\Big)$
$\Big(\dfrac{6}{7}\Big)$