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New answer posted

a year ago

0 Follower 6 Views

P
Payal Gupta

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar

Since,  the  two  given  events  are  not  related  to  the  same  Sample  space.∴The  sum  of  probabilities  of  two  students  getting  distinction  in  their  final      may  be  1.2Hence,  the  given  statement  is  'True'

New answer posted

a year ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar

H e r e ,     P ( A ) = 0 . 7 ,     P ( B ) = 0 . 3 ∴     P ( A ∩ B ) = P ( A ) * P ( B )                                                       = 0 . 7 * 0 . 3 = 0 . 2 1 B u t     t h e     g i v e n     p r o b a b i l i t y     i s     0 . 4 H e n c e ,     t h e     g i v e n     s t a t e m e n t     i s     ' F a l s e '

New answer posted

a year ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar

H e r e ,     P ( A ∩ B ) ≤ P ( A )     W h i c h     i s     a l w a y s     t r u e . H e n c e ,     t h e     g i v e n     s t a t e m e n t     i s     ' T r u e '

New answer posted

a year ago

0 Follower 4 Views

P
Payal Gupta

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar

G i v e n     t h a t     P ( A ) = 0 . 5     a n d     P ( A ∩ B ) ≤ 0 . 3 N o w     P ( A ) * P ( B ) ≤ 0 . 3 ⇒                           0 . 5 * P ( B ) ≤ 0 . 3 ⇒                                               P ( B ) ≤ 0 . 3 0 . 5 ⇒                                               P ( B ) ≤ 0 . 6 H e n c e ,     t h e     g i v e n     s t a t e m e n t     i s     ' F a l s e '

New answer posted

a year ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar

S u m     o f     a l l     p r o b a b i l i t i e s = 1 ∴ P ( 0 ) + P ( 1 ) + P ( 2 ) + P ( 3 ) + P ( 4 ) + P ( 5 ) = 0 . 1 2 + 0 . 2 5 + 0 . 3 6 + 0 . 1 4 + 0 . 0 8 + 0 . 1 1                                                                                                                                                                                       = 1 . 0 6 > 1 H e n c e ,     t h e     g i v e n     s t a t e m e n t     i s     ' F a l s e ' .

New answer posted

a year ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar

L e t     E     b e     t h e     e v e n t     t h a t     t h e     s t u d e n t     w i l l     p a s s a n d     F     b e     t h e     e v e n t     t h a t     h e     w i l l     g e t     c o m p a r t m e n t ∴ P ( E ) = 0 . 7 3 , P ( F ) = 0 . 1 3     a n d     P ( E ∪ F ) = 0 . 9 6 ∴                   P ( E ∪ F ) = P ( E ) + P ( F ) − P ( E ∩ F )                                                                     = 0 . 7 3 + 0 . 1 3 − 0                       [ ? P ( E ∩ F ) = 0 ]                                                                     = 0 . 8 6 B u t         P ( E ∪ F ) = 0 . 9 6 H e n c e ,     t h e     g i v e n     s t a t e m e n t     i s     ' F a l s e ' .

New answer posted

a year ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar

G i v e n     t h a t :                                                                   P ( t o     s e e     g i r a f f e e ) = 0 . 7 2                                                                               P ( t o     s e e     b e a r ) = 0 . 8 4 P ( t o     s e e     b o t h     g i r a f f e e     a n d     b e a r ) = 0 . 5 2 ∴                         P ( t o     s e e     g i r a f f e e     o r     b e a r ) = P ( t o     s e e     g i r a f f e e ) + P ( t o     s e e     b e a r ) − P ( t o     s e e     b o t h )                                                                                                                                                   = 0 . 7 2 + 0 . 8 4 − 0 . 5 2                                                                                                                                                   = 1 . 0 4     w h i c h     i s     n o t     p o s s i b l e . H e n c e ,     t h e     g i v e n     s t a t e m e n t     i s     ' F a l s e ' .

New answer posted

a year ago

0 Follower 5 Views

P
Payal Gupta

Contributor-Level 10

This is a matching answer type question as classified in NCERT Exemplar

( a ) I f     E 1     a n d     E 2     a r e     m u t u a l l y     e x c l u s i v e     e v e n t s ,     t h e n     E 1 ∩ E 2 = ? . ( b ) I f     E 1     a n d     E 2     a r e     m u t u a l l y     e x c l u s i v e     a n d     e x h a u s t i v e     e v e n t s ,     t h e n             E 1 ∩ E 2 = ?     a n d     E 1 ∪ E 2 = S . ( c ) I f     E 1     a n d     E 2     h a v e     c o m m o n     o u t c o m e s ,     t h e n     ( E 1 − E 2 ) ∪ ( E 1 ∩ E 2 ) = E 1 ( d ) I f     E 1     a n d     E 2     a r e     t w o     e v e n t s     s u c h     t h a t     E 1 ⊂ E 2         ⇒ E 1 ∩ E 2 = E 1               H e n c e ,     ( a ) ↔ ( i v ) , ( b ) ↔ ( i i i ) , ( c ) ↔ ( i i ) , ( d ) ↔ ( i )

New answer posted

a year ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

This is a matching answer type question as classified in NCERT Exemplar

( a ) 0 . 9 5 = V e r y     l i k e l y     t o     h a p p e n ,     s o     i t     i s     c l o s e     t o     1 . ( b ) 0 . 0 2 = V e r y     l i t t l e     c h a n c e     o f     h a p p e n i n g     a s     t h e     p r o b a b i l i t y     i s     v e r y     l o w . ( c ) − 0 . 3 = A n     i n c o r r e c t     a s s i g n m e n t     b e c a u s e     p r o b a b i l i t y     i s     n e v e r     n e g a t i v e . ( d )         0 . 5 = A s     m u c h     c h a n c e     o f     h a p p e n i n g     a s     n o t     b e c a u s e     s u m     o f     c h a n c e s     o f     h a p p e n i n g                                               a n d     n o t     h a p p e n i n g     i s     o n e . ( e )                   0 = N o     c h a n c e     o f     h a p p e n i n g .               H e n c e ,     ( a ) ↔ ( i v ) , ( b ) ↔ ( v ) , ( c ) ↔ ( i ) , ( d ) ↔ ( i i i ) , ( e ) ↔ ( i i )

New answer posted

a year ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

G i v e n     t h a t :                         P ( A ) = 0 . 5 ,     P ( B ) = 0 . 3 a n d                                       P ( A ∩ B ) = 0                 [ ? A     a n d     B     a r e     m u t u a l l y     e x c l u s i v e     e v e n t s ] ∴                                                   P ( A ¯ ∩ B ¯ ) = P ( A ∪ B ¯ )                                                                                                     = 1 − P ( A ∪ B ) = 1 − [ P ( A ) + P ( B ) ]                                                                                                     = 1 − ( 0 . 5 + 0 . 3 ) = 1 − 0 . 8 = 0 . 2 H e n c e ,     t h e     v a l u e     o f     t h e     f i l l e r     i s     0 . 2

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