Two distinct numbers are selected at random from the first twelve natural numbers. The probability that the sum will be divisible by 3 is

\frac{1}{3}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{23}{66}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

none of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{1}{3}$
$F i r s t 12 n a t u r a l n u m b e r s a r e 1, 2, 3, . . . . . . . ., 12 N u m b e r o f w a y s i n w h i c h t w o d i s t i n c t n u m b e r s c a n b e s e l e c t e d o u t o f t h e s e 12 n a t u r a l n u m b e r s = (\frac{12}{2}) L e t P (T) = (\frac{12}{2}) T h e m a x i m u m s u m o f t w o s e l e c t e d d i s t i n c t n u m b e r s = 12 + 11 = 23 T h e m i n i m u m s u m o f t w o s e l e c t e d d i s t i n c t n u m b e r s = 1 + 2 = 3 F o r t h e s u m b e i n g d i v i s i b l e b y 3 i t m u s t b e o n e o f 3, 6, 9, 12, 15, 18, 21 L e t S b e t h e e v e n t o f s u m o f t w o s e l e c t e d n u m b e r . P (S = 3) = \frac{1}{P (T)} P (S = 6) = \frac{2}{P (T)} P (S = 9) = \frac{4}{P (T)} P (S = 12) = \frac{5}{P (T)} P (S = 15) = \frac{5}{P (T)} P (S = 18) = \frac{3}{P (T)} P (S = 21) = \frac{2}{P (T)} L e t D b e t h e e v e n t t h a t s u m i s d i v i s i b l e b y 3 T h e n P (D) = P (S = 3) + P (S = 6) + P (S = 9) + P (S = 12) + P (S = 15) + P (S = 18) + P (S = 21) \Rightarrow P (D) = \frac{22}{P (T)} = \frac{22}{66} \Rightarrow P (D) = \frac{1}{3}$

Suggest Corrections

Similar questions

10

100

2

3

1, 2, 3, . . . . . . . . .100

3

a

b

30

a^{2} - b^{2}

3

150

3

5

Join BYJU'S Learning Program