MCQ
The probability that a leap year selected at random will have $53$ Fridays is :
  • $\frac{2}{7}$
  • B
    $\frac{6}{7}$
  • C
    $\frac{4}{7}$
  • D
    $\frac{1}{7}$

Answer

Correct option: A.
$\frac{2}{7}$
Leap year contains $366$ days $= 364$ days $+\ 2$ days $= 364/73$ weeks $+\ 2$ additional days $= 52$ weeks $+\ 2$ additional days
$52$ weeks contain $52$ Fridays
We will get $53$ Fridays if one of the remaining two additional days is a Friday
These additional days can be :
$\text{(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday)}$
$\text{(Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)}$
Number of total outcomes $= 7$
Number of possible outcomes $= 2$
$\therefore$ Required Probabillity of the event $=\frac{\text{Number of possible outcomes}}{\text{Number of total outcomes }}=\frac{2}{7}$

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