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A team of medical students doing their internship has to assist during surgeries at a city hospital. The probabilities of surgeries rated as very complex, complex, routine, simple, or very simple are respectively, 0.15, 0.20, 0.31, 0.26, 0.08. Find the probabilities that a particular surgery will be rated:

(a) complex or very complex;

(b) neither very complex nor very simple;

(c) routine or complex;

(d) routine or simple.

0 2 Views | Posted 3 months ago
Asked by Shiksha User

  • 1 Answer

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    Answered by

    Payal Gupta | Contributor-Level 10

    3 months ago

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

    L e t     E 1 , E 2 , E 3 , E 4     a n d     E 5     b e     t h e     e v e n t s     t h a t     t h e     s u r g e r i e s     a r e     r a t e d     a s     v e r y     c o m p l e x ,     c o m p l e x , r o u t i n e ,     s i m p l e     a n d     v e r y     s i m p l e     r e s p e c t i v e l y . ∴ P ( E 1 ) = 0 . 1 5 ,     P ( E 2 ) = 0 . 2 0 ,     P ( E 3 ) = 0 . 3 1 ,     P ( E 4 ) = 0 . 2 6     a n d     P ( E 5 ) = 0 . 0 8 ( a ) P ( c o m p l e x     a n d     v e r y     c o m p l e x ) = P ( E 1     o r     E 2 ) ⇒ P ( E 1 ∪ E 2 ) = P ( E 1 ) + P ( E 2 ) − P ( E 1 ∩ E 2 )                                                                 = 0 . 1 5 + 0 . 2 0 − 0                                                                 = 0 . 3 5                             [ ?     A l l     e v e n t     a r e     i n d e p e n d e n t ] ( b ) P ( n e i t h e r     v e r y     c o m p l e x     n o r     v e r y     s i m p l e ) = P ( E 1 ' ∩ E 5 ' )                                                                 = P ( E 1 ∪ E 5 ) ' = 1 − P ( E 1 ∪ E 5 )                                                                 = 1 − [ P ( E 1 ) + P ( E 5 ) ] = 1 − [ 0 . 1 5 + 0 . 0 8 ]                                                                 = 1 − 0 . 2 3 = 0 . 7 7 ( c ) P ( r o u t i n e     o r     c o m p l e x ) = P ( E 3     o r     E 2 ) ⇒ P ( E 3 ∪ E 2 ) = P ( E 3 ) + P ( E 2 )                                                                 = 0 . 3 1 + 0 . 2 0 = 0 . 5 1 ( d ) P ( r o u t i n e     o r     s i m p l e ) = P ( E 3     o r     E 4 ) ⇒ P ( E 3 ∪ E 4 ) = P ( E 3 ) + P ( E 4 )                                                                 = 0 . 3 1 + 0 . 2 6 = 0 . 5 7

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