Explanation Given: The probability P k , j that a mother cell with k plasmids of type R 1 produces a daughter cell with j plasmids of type R 1 can be calculated by the formula P k , j = ( 2 k j ) ( 4 − 2 k 2 − j ) ( 4 2 ) Assume that ( 0 0 ) = 1 and ( k j ) = 0 when k < j . We have an arithmetic Series. We use the sum of n numbers in Arithmetic Series formula. We take n t h term of Arithmetic Series and plug the value of a n in it. Formula used: Combination formula ( n r ) = C ( n , r ) = n ! ( n − r ) ! r ! , where n is the total number of objects and r is the number of objects selected. Calculation: Description P k , j = ( 2 k j ) ( 4 − 2 k 2 − j ) ( 4 2 ) and ( n r ) = C ( n , r ) = n ! ( n − r ) ! r ! Step1: Substitute the value of k and j in the equation P k , j to find P 00 P 00 = ( 2 ( 0 ) ( 0 ) ) ( 4 − 2 ( 0 ) 2 − ( 0 ) ) ( 4 2 ) = ( 0 0 ) ( 4 − 0 2 − 0 ) ( 4 2 ) = ( 0 0 ) ( 4 2 ) ( 4 2 ) ( 0 0 ) = 1 ( 4 2 ) = C ( 4 , 2 ) = 4 ! ( 4 − 2 ) ! 2 ! = 4 ! ( 2 ) ! 2 ! = 4 × 3 × 2 × 1 2 × 1 × 2 × 1 = 3 × 2 × 1 1 = 6 P 00 = ( 0 0 ) ( 4 2 ) ( 4 2 ) = 1 × 6 6 = 1 Step2: Substitute the value of k and j in the equation P k , j to find P 01 P 01 = ( 2 ( 0 ) ( 1 ) ) ( 4 − 2 ( 0 ) 2 − ( 1 ) ) ( 4 2 ) = ( 0 1 ) ( 4 − 0 2 − 1 ) ( 4 2 ) = ( 0 1 ) ( 4 1 ) ( 4 2 ) ( 0 1 ) = 0 since k < j ( 4 1 ) = C ( 4 , 1 ) = 4 ! ( 4 − 1 ) ! 1 ! = 4 ! ( 3 ) ! 1 ! = 4 × 3 × 2 × 1 3 × 2 × 1 × 1 = 4 1 = 4 ( 4 2 ) = C ( 4 , 2 ) = 4 ! ( 4 − 2 ) ! 2 ! = 4 ! ( 2 ) ! 2 ! = 4 × 3 × 2 × 1 2 × 1 × 2 × 1 = 3 × 2 × 1 1 = 6 P 01 = ( 0 1 ) ( 4 1 ) ( 4 2 ) = 0 × 4 6 = 0 Step3: Substitute the value of k and j in the equation P k , j to find P 02 P 02 = ( 2 ( 0 ) ( 2 ) ) ( 4 − 2 ( 0 ) 2 − ( 2 ) ) ( 4 2 ) = ( 0 2 ) ( 4 − 0 2 − 2 ) ( 4 2 ) = ( 0 2 ) ( 4 0 ) ( 4 2 ) ( 0 2 ) = 0 since k < j ( 4 0 ) = C ( 4 , 0 ) = 4 ! ( 4 − 0 ) ! 0 ! = 4 ! ( 4 ) ! 0 ! = 1 1 = 1 ( 4 2 ) = C ( 4 , 2 ) = 4 ! ( 4 − 2 ) ! 2 ! = 4 ! ( 2 ) ! 2 ! = 4 × 3 × 2 × 1 2 × 1 × 2 × 1 = 3 × 2 × 1 1 = 6 P 02 = ( 0 2 ) ( 4 0 ) ( 4 2 ) = 0 × 1 6 = 0 Step4: Substitute the value of k and j in the equation P k , j to find P 10 P 10 = ( 2 ( 1 ) ( 0 ) ) ( 4 − 2 ( 1 ) 2 − ( 0 ) ) ( 4 2 ) = ( 2 0 ) ( 4 − 2 2 − 0 ) ( 4 2 ) = ( 2 0 ) ( 2 2 ) ( 4 2 ) ( 2 0 ) = C ( 2 , 0 ) = 2 ! ( 2 − 0 ) ! 0 ! = 2 ! ( 2 ) ! 0 ! = 1 1 = 1 ( 2 2 ) = C ( 2 , 2 ) = 2 ! ( 2 − 2 ) ! 2 ! = 2 ! ( 0 ) ! 2 ! = 1 1 = 1 ( 4 2 ) = C ( 4 , 2 ) = 4 ! ( 4 − 2 ) ! 2 ! = 4 ! ( 2 ) ! 2 ! = 4 × 3 × 2 × 1 2 × 1 × 2 × 1 = 3 × 2 × 1 1 = 6 P 10 = ( 2 0 ) ( 2 2 ) ( 4 2 ) = 1 × 1 6 = 1 6 Step5: Substitute the value of k and j in the equation P k , j to find P 11 P 11 = ( 2 ( 1 ) ( 1 ) ) ( 4 − 2 ( 1 ) 2 − ( 1 ) ) ( 4 2 ) = ( 2 1 To determine b) The greatest element in P and to interpret the result.

5th Edition

ISBN: 9780321845092

Author: Rockswold

Publisher: YUZU

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Chapter 8, Problem 1EDE

To determine

a)

$P_{k, j}$ for $0 \leq k, j \leq 2$ and to record in the matrix $P = [\begin{matrix} P_{00} & P_{01} & P_{02} \\ P_{10} & P_{11} & P_{12} \\ P_{20} & P_{21} & P_{22} \end{matrix}]$ .

To determine

b)

The greatest element in P and to interpret the result.

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Chapter 8, Problem 77CRE

Chapter 8, Problem 2EDE

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Chapter 8 Solutions

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Ch. 8.1 - Prob. 1E Ch. 8.1 - Prob. 2E Ch. 8.1 - Prob. 3E Ch. 8.1 - Prob. 4E Ch. 8.1 - Prob. 5E Ch. 8.1 - Prob. 6E Ch. 8.1 - Prob. 7E Ch. 8.1 - Prob. 8E Ch. 8.1 - Prob. 9E Ch. 8.1 - Prob. 10E

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a) P k , j for 0 ≤ k , j ≤ 2 and to record in the matrix P = [ P 00 P 01 P 02 P 10 P 11 P 12 P 20 P 21 P 22 ] .

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a) Pk,j for 0≤k, j≤2 and to record in the matrix P=[P00P01P02P10P11P12P20P21P22] .

b) The greatest element in P and to interpret the result.

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Chapter 8 Solutions

a)

$P_{k, j}$ for $0 \leq k, j \leq 2$ and to record in the matrix $P = [\begin{matrix} P_{00} & P_{01} & P_{02} \\ P_{10} & P_{11} & P_{12} \\ P_{20} & P_{21} & P_{22} \end{matrix}]$ .

b)

The greatest element in P and to interpret the result.