概率统计历届考研汇编

( x > a ) 与\r0 , 其它\rB = ( y > a ) 独立， P(A + 6 ) = j , 求：( 1 ) 。值 (2) 的期望。\r’ 3\r解 ( 1 ) 由设 X 〜 /* ) = ( 京 2 ° < ” < 2 且 丫 与 x 同分布，\r0 , 其它\r4 = (X > a ) 与 B = ( y > a ) 独立，可知当a < 0 时\r+a> 0 2 Q +00 ]\rP(A) = P (X > a )= J /(x )d x = J ()d x + 卜 = —x 3 ^ - 1\ra a 0 8 2 8\r+ao\rP(B) = P ( Y > a ) = jf ( y ) d y = 1, B|J\ra\rP(A + B) = P (A ) + P(B) - P (A )P (B ) = 1 + 1 - 1 x 1 = 1 与\r3\r尸04 + 8 ) = )相 矛 盾 ，因而a 2 0 , 即\r+oo 2 Q +co [ [\rP(A) = P ( X 〉a )= j/ ( x ) J x = | - x 2J x + jOJx = - x 3 |^ = - ( 8 - a 3 )\ra a 2\r+O0 [\rP⑻ = P ( Y > a ) = J / ( y ) ^ = - ( 8 - a 3 ) , 即\rP(A + B) = P (4 ) + P (B )_ P (A )P (B )