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Probability question

2025 · 22 Jan · Shift 1 · Q35
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Probability question

2025 · 22 Jan · Shift 1 · Q35

JEE MainMathematicsProbabilityMCQ+4 / −1
A coin is tossed three times. Let XXX denote the number of times a tail follows a head. If μ\muμ and σ2\sigma^2σ2 denote the mean and variance of XXX, then the value of 64(μ+σ2)64\left(\mu+\sigma^2\right)64(μ+σ2) is:
  1. A
    64
  2. B
    32
  3. C
    51
  4. D
    48
View written solutionFree

Correct answer: D

  1. Define the random variable

A coin is tossed 3 times. Let XXX be the number of times a tail follows a head.

This means we check consecutive pairs:

  • toss 1 to toss 2
  • toss 2 to toss 3

For each pair, we count 1 if the pattern is HTHTHT.

So, X=I1+I2X = I_1 + I_2X=I1​+I2​ where

  • I1=1I_1 = 1I1​=1 if tosses (1,2)=HT(1,2) = HT(1,2)=HT, else 000
  • I2=1I_2 = 1I2​=1 if tosses (2,3)=HT(2,3) = HT(2,3)=HT, else 000
  1. List all outcomes and compute XXX

There are 23=82^3=823=8 equally likely outcomes:

OutcomeXHHH0HHT1(HH,HT)HTH1(HT,TH)HTT1(HT,TT)THH0THT1(TH,HT)TTH0TTT0\begin{array}{c|c} \text{Outcome} & X \\ \hline HHH & 0 \\ HHT & 1 \quad (HH, HT) \\ HTH & 1 \quad (HT, TH) \\ HTT & 1 \quad (HT, TT) \\ THH & 0 \\ THT & 1 \quad (TH, HT) \\ TTH & 0 \\ TTT & 0 \end{array}OutcomeHHHHHTHTHHTTTHHTHTTTHTTT​X01(HH,HT)1(HT,TH)1(HT,TT)01(TH,HT)00​​

Notice that X=2X=2X=2 is impossible, because if tosses (1,2)=HT(1,2)=HT(1,2)=HT, then toss 2 is TTT, so tosses (2,3)(2,3)(2,3) cannot also be HTHTHT.

Thus the distribution is:

  • P(X=0)=48=12P(X=0)=\frac{4}{8}=\frac12P(X=0)=84​=21​
  • P(X=1)=48=12P(X=1)=\frac{4}{8}=\frac12P(X=1)=84​=21​
  • P(X=2)=0P(X=2)=0P(X=2)=0
  1. Find the mean μ\muμ
μ=E[X]=0⋅12+1⋅12=12\mu = E[X] = 0\cdot \frac12 + 1\cdot \frac12 = \frac12μ=E[X]=0⋅21​+1⋅21​=21​

So, μ=12\mu = \frac12μ=21​

  1. Find E[X2]E[X^2]E[X2]
E[X2]=02⋅12+12⋅12=12E[X^2] = 0^2\cdot \frac12 + 1^2\cdot \frac12 = \frac12E[X2]=02⋅21​+12⋅21​=21​
  1. Find the variance σ2\sigma^2σ2
σ2=E[X2]−(E[X])2\sigma^2 = E[X^2] - (E[X])^2σ2=E[X2]−(E[X])2 σ2=12−(12)2=12−14=14\sigma^2 = \frac12 - \left(\frac12\right)^2 = \frac12 - \frac14 = \frac14σ2=21​−(21​)2=21​−41​=41​

So, σ2=14\sigma^2 = \frac14σ2=41​

  1. Compute 64(μ+σ2)64(\mu+\sigma^2)64(μ+σ2)
μ+σ2=12+14=34\mu+\sigma^2 = \frac12 + \frac14 = \frac34μ+σ2=21​+41​=43​

Therefore,

64(μ+σ2)=64⋅34=4864\left(\mu+\sigma^2\right)=64\cdot \frac34 = 4864(μ+σ2)=64⋅43​=48
  1. Compare with options

The correct option is:

48\boxed{48}48​

So the answer is Option D.

  1. Comparison with stored correct answer

Stored correct answer: D

Our derived answer: D

They agree.

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