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Probability question

2023 · 24 Jan · Shift 1 · Q25
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  5. /2023 · 24 Jan · Shift 1 · Q25

Probability question

2023 · 24 Jan · Shift 1 · Q25

JEE MainMathematicsProbabilityMCQ+4 / −1
Let N denote the number that turns up when a fair die is rolled. If the probability that the system of equations x+y+z=12x+Ny+2z=23x+3y+Nz=3x + y + z = 12x + \mathrm{N}y + 2z = 23x + 3y + \mathrm{N}z = 3x+y+z=12x+Ny+2z=23x+3y+Nz=3 has unique solution is k6{k \over 6}6k​, then the sum of value of k and all possible values of N is :
  1. A
    18
  2. B
    21
  3. C
    20
  4. D
    19
View written solutionFree

Correct answer: C

  1. Interpret the system correctly

The given system is

{x+y+z=12x+Ny+2z=23x+3y+Nz=3\begin{cases} x+y+z=1\\ 2x+Ny+2z=2\\ 3x+3y+Nz=3 \end{cases}⎩⎨⎧​x+y+z=12x+Ny+2z=23x+3y+Nz=3​

A system of 3 linear equations in 3 variables has a unique solution iff the determinant of its coefficient matrix is non-zero.

So we consider

A=(1112N233N)A=\begin{pmatrix} 1&1&1\\ 2&N&2\\ 3&3&N \end{pmatrix}A=​123​1N3​12N​​

and require

det⁡(A)≠0.\det(A)\neq 0.det(A)=0.
  1. Compute the determinant
det⁡(A)=∣1112N233N∣\det(A)=\begin{vmatrix} 1&1&1\\ 2&N&2\\ 3&3&N \end{vmatrix}det(A)=​123​1N3​12N​​

Expanding along the first row,

det⁡(A)=1∣N23N∣−1∣223N∣+1∣2N33∣.\det(A)=1\begin{vmatrix}N&2\\3&N\end{vmatrix}-1\begin{vmatrix}2&2\\3&N\end{vmatrix}+1\begin{vmatrix}2&N\\3&3\end{vmatrix}.det(A)=1​N3​2N​​−1​23​2N​​+1​23​N3​​.

Now,

∣N23N∣=N2−6,\begin{vmatrix}N&2\\3&N\end{vmatrix}=N^2-6,​N3​2N​​=N2−6, ∣223N∣=2N−6,\begin{vmatrix}2&2\\3&N\end{vmatrix}=2N-6,​23​2N​​=2N−6, ∣2N33∣=6−3N.\begin{vmatrix}2&N\\3&3\end{vmatrix}=6-3N.​23​N3​​=6−3N.

Therefore,

det⁡(A)=(N2−6)−(2N−6)+(6−3N).\det(A)=(N^2-6)-(2N-6)+(6-3N).det(A)=(N2−6)−(2N−6)+(6−3N).

Simplifying,

det⁡(A)=N2−5N+6=(N−2)(N−3).\det(A)=N^2-5N+6=(N-2)(N-3).det(A)=N2−5N+6=(N−2)(N−3).
  1. Find values of NNN for which unique solution exists

A fair die can show

N∈{1,2,3,4,5,6}.N\in\{1,2,3,4,5,6\}.N∈{1,2,3,4,5,6}.

Unique solution exists when

(N−2)(N−3)≠0,(N-2)(N-3)\neq 0,(N−2)(N−3)=0,

that is,

N≠2,3.N\neq 2,3.N=2,3.

Hence possible values of NNN giving a unique solution are

1,4,5,6.1,4,5,6.1,4,5,6.

So the probability is

\frac{4}{6}= rac{k}{6} \implies k=4.
  1. Compute the required sum

“All possible values of NNN” here means the values of NNN for which the system has a unique solution, namely

1,4,5,6.1,4,5,6.1,4,5,6.

Their sum is

1+4+5+6=16.1+4+5+6=16.1+4+5+6=16.

So,

k+(sum of all possible values of N)=4+16=20.k+\text{(sum of all possible values of }N)=4+16=20.k+(sum of all possible values of N)=4+16=20.
  1. Final answer
20\boxed{20}20​

So the correct option is C.

  1. Comparison with stored answer

Stored correct answer: C

Our derived answer is also C.

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