Magdy Essafty – [email protected] 1 الرد على موضوع سؤال فى الحتمالت

اسم العضو : مجدى الصفتى

Question 1 If A and B are independent events Then P ( A ∩B ) = P ( A ) . P ( B ) To prove that : A, B  are independent events we must prove : P ( A ∩B ) = P ( A) . P ( B ) Since : P ( A ∩B ) = 1 – P ( A B ) P ( A ∩B ) = 1 – P ( A ) – P ( B ) + P ( A ∩B ) P ( A ∩B ) = 1 – P ( A ) – P ( B ) + P ( A ) . P ( B ) P ( A ∩B ) = [ 1 – P ( A ) ] – P ( B ) [ 1 – P ( A ) ] P ( A ∩B ) = [ 1 – P ( A ) ] . [ 1 – P ( B ) ] P ( A ∩B ) = P ( A) . P ( B ) Therefore A, B  are independent events.

Question 2 S = { ( w , w , w , w ) , ( l , l , l , l ) , ( w , w , w , l ) , ( w , w , l , w ) , ( w , l , w , w ) , ( l , w , w , w ) , ( l , l , l , w ) , ( l , l , w , l ) , ( l , w , l , l ) , ( w , l , l , l ) , ( w , w , l , l ) , ( l , l , w , w ) , ( w , l , l , w ) , ( l , w , w , l ) , ( w , l , w , l ) , ( l , w , l , w ) } Range of the random variable: { 4 , 3 , 2 , 1 , 0 } The probability distribution:

X r 4 3 2 1 0 1 1 3 1 1 f ( X r ) 16 4 8 4 16 Magdy Essafty – [email protected] 2 الرد على موضوع سؤال فى الحتمالت

اسم العضو : مجدى الصفتى

The expectation 1 1 3 1 1  = 4  + 3  + 2  + 1  + 0  = 2 16 4 8 4 16 The variance 1 1 3 1 1  2 = 16  + 9  + 4  + 1  + 0  = 5 16 4 8 4 16

Question 3

 = 250 ,  = 6 240 250 P ( x  240 ) = P ( z  ) 6

P ( z  - 1.67 ) = P ( z  1.67 ) = 0.5 – 0.4525 = 0.0475

Question 4

ىىى ىىى ىىىىى ىىىىىىىى ىى ىىىىىى ىىىىىى :

10  7  3 C 3 C 4 C 3 = 4200

ىىى ىىى ىىىىى ىىىىىىىى ىى ىىىىىى ىىىىىى :

3  7  3 C 3 C 4 C 3 = 35 مع تحياتى للجميع