what is the probability that a non-leap year selected randomly will (i) have 53 Sundays and (ii) 52 Sundays

Open in App

Solution

Dear Student,
$A n o n - l e a p y e a r h a s 365 d a y s A y e a r h a s 52 w e e k s . H e n c e t h e r e w i l l b e 52 S u n d a y s f o r s u r e . 52 w e e k s = 52 \times 7 = 364 d a y s . 365 - 364 = 1 d a y e x t r a . I n a n o n - l e a p y e a r t h e r e w i l l b e 52 S u n d a y s a n d 1 d a y w i l l b e l e f t . T h i s 1 d a y c a n b e S u n d a y, M o n d a y, T u e s d a y, W e d n e s d a y, T h u r s d a y, f r i d a y, S a t u r d a y, S u n d a y . O f t h e s e t o t a l 7 o u t c o m e s, t h e f a v o u r a b l e o u t c o m e s a r e 1 . H e n c e t h e p r o b a b i l i t y o f g e t t i n g 53 s u n d a y s = \frac{1}{7} 2) ∴ p r o b a b i l i t y o f g e t t i n g 52 s u n d a y s = 1 - \frac{1}{7} = \frac{6}{7}$
Regards

Suggest Corrections

Similar questions

53

53

53

Join BYJU'S Learning Program