if-three-numbers-are-drawn-at-random-successively-without-replacement-from-a-set-S-1-2-10-then-probability-that-the-minimum-of-the-choosen-number-is-3-or-their-maximum-is-7-Answer-11

Question Number 40550 by vajpaithegrate@gmail.com last updated on 24/Jul/18

if three numbers are drawn at random

successively without replacement from

a set S = {1, 2, \dots \dots 10} then probability that

the minimum of the choosen number is

3 or their maximum is 7

Answer = \frac{11}{40}

Answered by tanmay.chaudhury50@gmail.com last updated on 24/Jul/18

P (A) = p r o b a b l i t y t o c h o o s e t h r e e n u m b e r

w h e r e 3 i s m i n i m u m

s o c h o o s i n g 3 a n d a n y t w o n u m b e r f r o m (4, 5,

6, 7, 8, 9, 10)

P (A) \frac{1 c_{1} \times 7 c_{2}}{10 c_{3}} = \frac{1 \times \frac{7 \times 6}{2}}{\frac{10 \times 9 \times 8}{3 \times 2}} = \frac{21}{120} = \frac{7}{40}

P (B) = p r o b a b l i g y t o c h o o s e t h r e e n u m b e r

w h e r e 7 i s m a x i m u m

s o c h o o s i n g 7 a n d a n y t w o n u m b e r f r o m

(1, 2, 3, 4, 5, 6)

P (B) \frac{1 c_{1} \times 6 c_{2}}{10 c_{3}} = \frac{15}{\frac{10 \times 9 \times 8}{3 \times 2}} = \frac{15}{120} = \frac{5}{40}

P (A) o r P (B)

P (A) \cup P (B) = P (A) + P (B) - P (A) \cap P (B)

c a l c u l a i o n o f P (A) \cap P (B)

(3, 4, 5, 6, 7) c h o o s i n g 3 a n d 7 a n d a n y o n e f r o m

(4, 5, 6)

P (A) \cap P (B) = \frac{1 c_{1} \times 1 c_{1} \times 3 c_{1}}{10 c_{3}} = \frac{3}{120} = \frac{1}{40}

s o r e q u i r e d r e s u l t s = \frac{7}{40} + \frac{5}{40} - \frac{1}{40} = \frac{11}{40}

Commented by vajpaithegrate@gmail.com last updated on 25/Jul/18

tnq sir