{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw7_418_10

hw7_418_10 - 48 Let X be the number of defective disks in...

This preview shows pages 1–2. Sign up to view the full content.

48. Let X be the number of defective disks in one package. X B (10 , 0 . 01). Then P ( X 1) = 0 . 99 10 + ( 10 1 ) 0 . 01(1 - 0 . 01) 9 = 0 . 09135. i.e. p ( X > 1) = 0 . 004248 . Let Y denotes the number of packages returned out of 3. Then Y B (3 , 0 . 004248). Then p ( Y = 1) = ( 3 1 ) 0 . 004248(1 - 0 . 004248) 2 = 0 . 01264 . 49. Let X1 denotes the number of times coin 1 lands on heads. X2 denotes the number of times coin 2 lands on heads. a) X 1 B (10 , 0 . 4), X 2 B (10 , 0 . 7). p ( X 1 = 7) = ( 10 7 ) 0 . 4 7 (1 - 0 . 4) 0 . 3 = 0 . 4247 , p ( X 2 = 7) = ( 10 7 ) 0 . 7 7 (1 - 0 . 7) 0 . 3 = 0 . 26683 . p(Coin lands on head exactly 7 of the 10 times)=p(X1=7)p(coin 1)+p(X2=7)p(coin 2)=0.0424(1/2)+0.26683*(1/2)=0.15465. b) A=coin 1 is selected. B=heads results, A C =coin 2 is selected. p ( A | B ) = p ( B | A ) p ( A ) p ( B | A ) p ( A )+ p ( B | A C ) p ( A C ) = 0 . 4 * (1 / 2) 0 . 4 * 1 / 2+0 . 7 * 1 / 2 . Given that the first ﬂip lands on heads, there are 9 tosses left and exactly 6 more need to land on heads. p ( X = 6) = p ( X = 6 | A ) p ( A ) + p ( X = 6 | A C ) p ( A C ) = 0 . 1968 . p ( X 1 = 6) = p ( X = 6 | A ) = ( 9 6 ) 0 . 4 6 (1 - 0 . 4) 3 = 0 . 743, p ( X 2 = 6) = p ( X = 6 | A C ) = ( 9 6 ) 0 . 7 6 (1 - 0 . 7) 3 = 0 . 2668. 51. X is the number of typographical errors on a page of a certain magazine.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 3

hw7_418_10 - 48 Let X be the number of defective disks in...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online