Reserve Problems Chapter 2 Section 3 Problem 7 Code 39 is a common bar code syst

Question

Reserve Problems Chapter 2 Section 3 Problem 7 Code 39 is a common bar code system that consists of narrow and wide bars (black) separated by either wide or narrow spaces (white). Each character contains nine elements (five bars and four spaces). The code for a character starts and ends with a bar (either narrow or wide) and a (white) space appears between each bar. The original specification (since revised) used exactly two wide bars and one wide space in each character. For example, if b and B denote narrow and wide (black) bars, respectively, and w and W denote narrow and wide (white) spaces, a valid character is bwBwBWbwb (the number 6). Suppose that all 40 codes are equally likely (none is held back as a delimiter). Determine the probability for each of the following (a) A wide space occurs before a narrow space (b) Two wide bars occur consecutively. (c) Two consecutive wide bars are at the start or end (d) The middle bar is wide.

Explanation / Answer

bar code has 9 elements 5 bar and 4 space

5 bar has 2 wide and 3 narrow bars

4 space has 1 wide and 3 narrow bar

total no. of ways of arranging these 9 elements=ways of arranging bars*ways of arranging space

ways of arranging bars=5!/(2!*3!) =10

ways of arranging space=(4!/3!) =4

total no. of ways of arranging these 9 elements=10*4=40 ways

P(bars)=10/40

P(space)=4/40

a)A wide space occurs before a narrow space

now, whatever be the postion of bars ,wide space will always be at second place,so 3 narrrow space can be arranged in 1 way

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so,P(A wide space occurs before a narrow space)=1*P(bars)=1*10/40=1/4

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b)no. of ways by which two wide bars occur simulataneously=BBbbb,bBBbb,bbBBb,bbbBB

no. of ways of two wide bars simulataneously=4!/3! * ways of arranging space

=4*4=16

so, P(two wide bars occur simulataneously)=16/40=0.4

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c)no. of ways of two wide bars occur at the start or end=BBbbb,bbbBB

no. of ways of two wide bars occur at the start or end=2 *ways of space

=2*4=8

P( two wide bars occur at the start or end)=8/40=0.2

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d)no. of codes having middle bar wide=bbBBb,bbBbB,BbBbb,bBBbb

no. of codes having middle bar wide=4!/3! * 4=16

P(middle bar wide)=16/40=0.4

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