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Probability question

2021 · 26 Aug · Shift 2 · Q31
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  5. /2021 · 26 Aug · Shift 2 · Q31

Probability question

2021 · 26 Aug · Shift 2 · Q31

JEE MainMathematicsProbabilityMCQ+4 / −1
Two fair dice are thrown. The numbers on them are taken as λ\lambdaλ and μ\muμ, and a system of linear equations x + y + z = 5 x + 2y + 3z = μ\muμ x + 3y +λ\lambdaλ z = 1 is constructed. If p is the probability that the system has a unique solution and q is the probability that the system has no solution, then :
  1. A
    p=16p = {1 \over 6}p=61​ and q=136q = {1 \over 36}q=361​
  2. B
    p=56p = {5 \over 6}p=65​ and q=536q = {5 \over 36}q=365​
  3. C
    p=56p = {5 \over 6}p=65​ and q=136q = {1 \over 36}q=361​
  4. D
    p=16p = {1 \over 6}p=61​ and q=536q = {5 \over 36}q=365​
View written solutionFree

Correct answer: B

  1. Write the system in matrix form

The system is

{x+y+z=5x+2y+3z=μx+3y+λz=1\begin{cases} x+y+z=5\\ x+2y+3z=\mu\\ x+3y+\lambda z=1 \end{cases}⎩⎨⎧​x+y+z=5x+2y+3z=μx+3y+λz=1​

Its coefficient matrix is

A=(11112313λ).A=\begin{pmatrix} 1&1&1\\ 1&2&3\\ 1&3&\lambda \end{pmatrix}.A=​111​123​13λ​​.

A system of 3 linear equations has a unique solution iff

det⁡(A)≠0.\det(A)\ne 0.det(A)=0.

If

det⁡(A)=0,\det(A)=0,det(A)=0,

then the system may have either no solution or infinitely many solutions.

  1. Compute the determinant
det⁡(A)=∣11112313λ∣\det(A)=\begin{vmatrix} 1&1&1\\ 1&2&3\\ 1&3&\lambda \end{vmatrix}det(A)=​111​123​13λ​​

Apply row operations:

R2→R2−R1,R3→R3−R1R_2\to R_2-R_1,\qquad R_3\to R_3-R_1R2​→R2​−R1​,R3​→R3​−R1​

Then

det⁡(A)=∣11101202λ−1∣\det(A)=\begin{vmatrix} 1&1&1\\ 0&1&2\\ 0&2&\lambda-1 \end{vmatrix}det(A)=​100​112​12λ−1​​

Expanding along the first column,

det⁡(A)=∣122λ−1∣=(λ−1)−4=λ−5.\det(A)=\begin{vmatrix} 1&2\\ 2&\lambda-1 \end{vmatrix} =(\lambda-1)-4=\lambda-5.det(A)=​12​2λ−1​​=(λ−1)−4=λ−5.

So:

  • unique solution when λ≠5\lambda\ne 5λ=5
  • not unique when λ=5\lambda=5λ=5

Since λ\lambdaλ is the outcome on a fair die, it can be any of 1,2,3,4,5,61,2,3,4,5,61,2,3,4,5,6 with equal probability.

Therefore,

p=P(λ≠5)=56.p=P(\lambda\ne 5)=\frac{5}{6}.p=P(λ=5)=65​.
  1. Now find when there is no solution

This can happen only when λ=5\lambda=5λ=5.

So substitute λ=5\lambda=5λ=5 into the system:

{x+y+z=5x+2y+3z=μx+3y+5z=1\begin{cases} x+y+z=5\\ x+2y+3z=\mu\\ x+3y+5z=1 \end{cases}⎩⎨⎧​x+y+z=5x+2y+3z=μx+3y+5z=1​

Notice that the third equation's left side is

x+3y+5z=2(x+2y+3z)−(x+y+z).x+3y+5z = 2(x+2y+3z)-(x+y+z).x+3y+5z=2(x+2y+3z)−(x+y+z).

Indeed,

2(x+2y+3z)−(x+y+z)=x+3y+5z.2(x+2y+3z)-(x+y+z)=x+3y+5z.2(x+2y+3z)−(x+y+z)=x+3y+5z.

So for consistency, the right sides must also satisfy the same relation:

1=2μ−5.1 = 2\mu - 5.1=2μ−5.

Hence

2μ=6  ⟹  μ=3.2\mu=6\implies \mu=3.2μ=6⟹μ=3.

Therefore:

  • if λ=5\lambda=5λ=5 and μ=3\mu=3μ=3, the system is consistent and has infinitely many solutions
  • if λ=5\lambda=5λ=5 and μ≠3\mu\ne 3μ=3, the system is inconsistent and has no solution

So the probability of no solution is

q=P(λ=5, μ≠3).q=P(\lambda=5,\ \mu\ne 3).q=P(λ=5, μ=3).

There are 363636 equally likely pairs (λ,μ)(\lambda,\mu)(λ,μ). For no solution:

  • λ=5\lambda=5λ=5 gives 1 choice
  • μ≠3\mu\ne 3μ=3 gives 5 choices

Hence,

q=536.q=\frac{5}{36}.q=365​.
  1. Match with the options

We found

p=56,q=536.p=\frac{5}{6},\qquad q=\frac{5}{36}.p=65​,q=365​.

This matches Option B.

  1. Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

So the stored answer is correct.

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