I-have-five-envelopes-numbered-3-4-5-6-7-all-hidden-in-a-box-i-picked-an-envelope-If-its-prime-then-i-get-the-square-of-that-number-in-Naira-Otherwise-without-replacement-i-picked-another-env

Question Number 7168 by Tawakalitu. last updated on 14/Aug/16

I h a v e f i v e e n v e l o p e s n u m b e r e d 3, 4, 5, 6, 7 a l l h i d d e n i n a

b o x, i p i c k e d a n e n v e l o p e . I f i t s p r i m e t h e n i g e t t h e

s q u a r e o f t h a t n u m b e r i n N a i r a . O t h e r w i s e (w i t h o u t

r e p l a c e m e n t) i p i c k e d a n o t h e r e n v e l o p e a n d t h e n g e t t h e

s u m o f t h e s q u a r e s o f t h e t w o n u m b e r s p i c k e d (i n N a i r a)

w h a t i s t h e c h a n c e o f m e g e t t i n g N 25 ?

Commented by Tawakalitu. last updated on 14/Aug/16

Y e s c o r r e c t \dots . . T h a n k y o u s o m u c h .

Commented by Rasheed Soomro last updated on 14/Aug/16

I c a n g e t N 25 o n l y i n t h e f o l l o w i n g t w o c a s e s :

(i) I f I p i c k e d 5 [∵ 5^{2} = 25]

(i i) I f I p i c k e d 4 a n d 3 i n s u c c e s s i o n [∵ 4^{2} + 3^{2} = 25]

I n c a s e (i) t h e c h a n c e o f g e t t i n g 5 i s \frac{1}{5}

I n c a s e (i i) p i c k i n g 4 h a s c h a n c e \frac{1}{5} a n d a f t e r

p i c k i n g 4, p i c k i n g 3 h a s \frac{1}{4} p r o b a b l i t y .

T h e r e f o r e p i c k i n g 4 a n d 3 i n s u c c e s s i o n h a s

p r o b a b l i t y \frac{1}{5} \times \frac{1}{4} = \frac{1}{20}

P r o b a b l i t y o f (p i c k i n g 5) o r (4 a n d 3 i n s u c c e s s i o n)

w i l l b e \frac{1}{5} + \frac{1}{20} = \frac{4 + 1}{20} = \frac{5}{20} = \frac{1}{4} ? ? ?