Question

what is the probability that a leap year selected at random will contain 53 Saturdays. [Hint: 366 = 52 × 7 + 2].​

Answer 1

Answer:

2/7 = 0.28 is probability for 53 Saturdays in a leap year.

Step-by-step explanation:

A normal year has  

• 52 Mondays
• 52 Tuesdays
• 52 Wednesdays
• 52 Thursdays
• 52 Fridays
• 52 Saturdays
• 52 Sundays  
• + 1 day that could be anything depending upon the year under consideration.  

In addition to this, a leap year has an extra day which might be a Monday or Tuesday or Wednesday or Sunday.

Our sample space is S : {Monday-Tuesday, Tuesday-Wednesday, Wednesday-Thursday,..., Sunday-Monday}

Number of elements in S = n(S) = 7

What we want is a set A (say) that comprises of the elements Saturday-Sunday and Sunday-Monday i.e. A : {Saturday-Sunday, Sunday-Monday}

Number of elements in set A = n(A) = 2

By definition, probability of occurrence of A = n(A)/n(S) = 2/7  

Therefore, probability that a leap year has 53 Sundays is 2/7.

$\dag{\boxed {\underline{\sf{\underline\orange{hope \: it \: helps \: u }}}}}\\\\\dag{\boxed {\underline{\sf{\underline\orange{Please \: mark \: as \: BRAINLIEST! }}}}}$

Answer 2

Answer:

Leap year = 366 days,

Weeks = 52,

Days = 2.

S = {(Sunday, Monday), (Monday, Tuesday),( Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday) }.

n(S) = 7

A---->> SATURDAY

A = {(Friday, Saturday), (Saturday, Sunday)}.

n(A) = 2

P(A) = n(A) /n(S) = 2/7.

Step-by-step explanation:

#Hope you have satisfied with this answer.