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JEE Mathematics 2023 Question with Solution

A coin is biased so that the head is 33 times as likely to occur as tail. This coin is tossed until a head or 33 tails occur. If XX denotes the number of tosses of the coin, then the mean of XX is:

  • A

    2116\frac{21}{16}

  • B

    1516\frac{15}{16}

  • C

    8164\frac{81}{64}

  • D

    3716\frac{37}{16}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The coin is biased with probability of head and tail as

P(H)=34,P(T)=14P(\text{H}) = \frac{3}{4}, \qquad P(\text{T}) = \frac{1}{4}

Find: The mean value of XX, where XX is the number of tosses until a head occurs or 33 tails occur.

Since the process stops as soon as the first head appears or when 33 tails have occurred, the possible values of XX are 1,2,31, 2, 3.

For X=1X = 1, the first toss must be a head:

P(X=1)=34P(X=1) = \frac{3}{4}

For X=2X = 2, the first toss must be a tail and the second toss must be a head:

P(X=2)=P(T)⋅P(H)=14×34=316P(X=2) = P(\text{T}) \cdot P(\text{H}) = \frac{1}{4} \times \frac{3}{4} = \frac{3}{16}

For X=3X = 3, the first two tosses must be tails, and then the third toss may be either head or tail:

P(X=3)=P(TTH)+P(TTT)P(X=3) = P(\text{TTH}) + P(\text{TTT}) =(14)2⋅34+(14)3= \left(\frac{1}{4}\right)^2 \cdot \frac{3}{4} + \left(\frac{1}{4}\right)^3 =364+164=464=116= \frac{3}{64} + \frac{1}{64} = \frac{4}{64} = \frac{1}{16}

Now compute the expectation:

E[X]=1⋅34+2⋅316+3⋅116\mathbb{E}[X] = 1 \cdot \frac{3}{4} + 2 \cdot \frac{3}{16} + 3 \cdot \frac{1}{16} =34+616+316= \frac{3}{4} + \frac{6}{16} + \frac{3}{16} =1216+916=2116= \frac{12}{16} + \frac{9}{16} = \frac{21}{16}

Therefore, the mean of XX is 2116\frac{21}{16}, so the correct option is A.

Case-wise Probability Listing

Given: Stopping occurs at the first head or after 33 tails in total.

Find: The expected number of tosses.

List all stopping sequences:

  • H\text{H} gives X=1X=1
  • TH\text{TH} gives X=2X=2
  • TTH\text{TTH} gives X=3X=3
  • TTT\text{TTT} gives X=3X=3

Their probabilities are:

P(H)=34,P(TH)=14⋅34=316P(\text{H}) = \frac{3}{4}, \qquad P(\text{TH}) = \frac{1}{4}\cdot\frac{3}{4} = \frac{3}{16} P(TTH)=(14)2⋅34=364,P(TTT)=(14)3=164P(\text{TTH}) = \left(\frac{1}{4}\right)^2\cdot\frac{3}{4} = \frac{3}{64}, \qquad P(\text{TTT}) = \left(\frac{1}{4}\right)^3 = \frac{1}{64}

Hence,

P(X=3)=364+164=116P(X=3) = \frac{3}{64} + \frac{1}{64} = \frac{1}{16}

and therefore

E[X]=1⋅34+2⋅316+3⋅116=2116\mathbb{E}[X] = 1\cdot\frac{3}{4} + 2\cdot\frac{3}{16} + 3\cdot\frac{1}{16} = \frac{21}{16}

Thus the required mean is 2116\frac{21}{16}.

Common mistakes

  • A common mistake is to take P(H)=33+1P(\text{H}) = \frac{3}{3+1} incorrectly as something other than 34\frac{3}{4}. Since head is 33 times as likely as tail, the ratio is 3:13:1, so the correct probabilities are 34\frac{3}{4} and 14\frac{1}{4}.

  • Students often miss that X=3X=3 includes both sequences TTH\text{TTH} and TTT\text{TTT}. It is wrong to count only one of them. The process stops on the third toss in either case.

  • Another mistake is to assume the experiment can continue beyond 33 tosses. That is not possible here because after 33 tails the stopping condition is already met.

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