Question

A, B, C in order to toss a coin. The first one to throw a head wins. If A starts to toss, then:
(a) \[P\left( A \right)=\dfrac{4}{7}\]
(b) \[P\left( B \right)=\dfrac{2}{7}\]
(c) \[P\left( C \right)=\dfrac{1}{7}\]
(d) \[P\left( C \right)=\dfrac{2}{7}\]

Answer 1

Hint: Just follow the logic for 1 to win the other 2 must lose in the game. As the game is starting with A we need A, B to lose in order to make C win.

Complete step-by-step answer:

First we will find the probability for A to win. For A to win the possible cases are:

(i)A wins directly, if he tosses the first toss into a head.

(ii)The next case for A winning is: first he has to lose in the first toss after that B, C must lose in order. Then in 4th toss A should toss head in order to win.

(iii)The next case is repetition of case (ii) then A in 7th toss A should toss head in order to win.

And so on infinite cases.

Let the probability of A winning is P(A).

P(A) = p(i) + p(ii) + p(iii)…..

Now we need to find p(i).

The case of falling head in first toss the probability is:

Tosses: H

\[p\left( i \right)=\dfrac{1}{2}\]

Now we need to find p(ii).

In this case A, B, and C should lose in the first round.

The case of falling head in 4th toss the probability is:

Tosses: TTTH

\[p\left( ii \right)=\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}\]

Now we need to find p(iii).

In this case A, B, and C should lose in the first and second rounds.

The case of falling head in 7th toss the probability is:

Tosses: TTTTTTH

\[p\left(
 iii \right)=\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}

\]

Now substituting these values into:

P(A) = p(i) + p(ii) + p(iii)…..

\[P\left( A \right)=\dfrac{1}{2}+{{\left( \dfrac{1}{2} \right)}^{4}}+{{\left( \dfrac{1}{2} \right)}^{7}}+...\]

We can see that this is a geometric progression.

Sum of infinite terms in a Geometric Progression with first term a and common ratio r is given by:

\[{{S}_{\infty }}=\dfrac{a}{1-r}\]

Here, in our case:

\[a=\dfrac{1}{2}\text{ and }r=\dfrac{1}{8}\]

By using the sum of G.P. formula:

\[P\left( A \right)=\dfrac{\dfrac{1}{2}}{1-\dfrac{1}{8}}=\dfrac{\dfrac{1}{2}}{\dfrac{7}{8}}\]
\[\]

Next we will find the probability for B to win. For B to win the possible cases are:

(i)B wins directly, if A tosses the first toss into a tail and next B tosses it as head.
(ii)The next case for B winning is: first A has to lose in the first toss after that B, C must lose in order. Then in 4th toss A should lose again and in 5th B must toss head in order to win.

(iii)The next case is repetition of case (ii) then B in 8th toss B should toss head in order to win.

And so on infinite cases.

Let the probability of B winning is P(B).

P(A) = p(i) + p(ii) + p(iii)…..

Now we need to find p(i).

The case of falling head in 2nd toss the probability is:

Tosses: TH

\[p\left( i \right)=\dfrac{1}{2}.\dfrac{1}{2}\]

Now we need to find p(ii).

In this case A, B, and C should lose in the first round.

The case of falling head in 5th toss the probability is:

Tosses: TTTTH

\[p\left( ii \right)=\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}\]

Now we need to find p(iii).

In this case A, B, and C should lose in the first and second rounds.

The case of falling head in 8th toss the probability is:

Tosses: TTTTTTTH

\[p\left(
 iii \right)=\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}
.\dfrac{1}{2}\]

Now substituting these values into:

P(B) = p(i) + p(ii) + p(iii)…..

\[P\left( B \right)={{\left( \dfrac{1}{2} \right)}^{2}}+{{\left( \dfrac{1}{2} \right)}^{5}}+{{\left(

\dfrac{1}{2} \right)}^{8}}+...\]

We can see that this is a geometric progression.

Sum of infinite terms in a Geometric Progression with first term a and common ratio r is given by:

\[{{S}_{\infty }}=\dfrac{a}{1-r}\]

Here, in our case:

\[a=\dfrac{1}{4}\text{ and }r=\dfrac{1}{8}\] By using the sum of G.P. formula:

\[P\left( B \right)=\dfrac{\dfrac{1}{4}}{1-\dfrac{1}{8}}=\dfrac{\dfrac{1}{4}}{\dfrac{7}{8}}\]

\[\]

Next we will find the probability for C to win. For C to win the possible cases are:

(i)C wins directly, if A, B tosses the first, second tosses as tails and next C tosses it as head.

(ii)The next case for C winning is: first A has to lose in first toss after that B, C must lose in

 order. Then in 4th and 5th tosses A and B must toss tail and lose and then in 6th toss C

should toss a head in order to win.

(iii)The next case is repetition of case (ii) then B in 9th toss B should toss head in order to win.

And so on infinite cases.

Let the probability of C winning is P(C).
P(C) = p(i) + p(ii) + p(iii)…..

Now we need to find p(i).

The case of falling head in 3rd toss the probability is:

Tosses: TTH

\[p\left( i \right)=\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}\]

Now we need to find p(ii).

In this case A, B, and C should lose in the first round.

The case of falling head in 6th toss the probability is:

Tosses: TTTTTH

\[p\left( ii \right)=\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}\]
Now we need to find p(iii).

In this case A, B, and C should lose in the first and second rounds.

The case of falling head in 9th toss the probability is:

Tosses: TTTTTTTTH


\[p\left( iii \right)=\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}

.\dfrac{1}{2}.\dfrac{1}{2}\]

Now substituting these values into:

P(B) = p(i) + p(ii) + p(iii)…..

\[P\left( C \right)={{\left( \dfrac{1}{2} \right)}^{3}}+{{\left( \dfrac{1}{2} \right)}^{6}}+{{\left(

\dfrac{1}{2} \right)}^{9}}+...\]

We can see that this is a geometric progression.

Sum of infinite terms in a Geometric Progression with first term a and common ratio r is

given by:

\[{{S}_{\infty }}=\dfrac{a}{1-r}\]

Here, in our case:

\[a=\dfrac{1}{8}\text{ and }r=\dfrac{1}{8}\]

By using sum of G.P. formula:

\[P\left( C \right)=\dfrac{\dfrac{1}{8}}{1-\dfrac{1}{8}}=\dfrac{\dfrac{1}{8}}{\dfrac{7}{8}}\]
\[\]

Therefore \[P\left( A \right)=\dfrac{4}{7},P\left( B \right)=\dfrac{2}{7},P\left( C \right)=\dfrac{1}{7}\].

So the options (a), (b), (c) are correct.

Note: Don’t forget to take the probability of toss which is head. Generally students take all probabilities but don’t take the probability of head as it is a winning toss. But that is wrong.