Question

In a certain race there are three boys a,b and c the winning probability of a is twice that of b and the winning probability of b is twice that of c if p(a)+p(b)+p(c)=1 then find the probability of their winning

Answer 1

Given:

Three boys a, b and c in a race with winning probabilities as:

Winning probability of a = 2 $\times$ Winning probability of b

And, Winning probability of b = 2 $\times$ Winning probability of c

Also, P(a) + P(b) + P(c) = 1

To Find:

The values of P(a), P(b) and P(c).

Solution:

Let, Winning probability of a = P(a)

Winning probability of b = P(b)

And, Winning probability of c = P(c).

Then, according to the given condition:

P(a) = 2 P(b)  .....(1)

P(b) = 2 P(c)  .....(2)

Putting the value of P(b) from equation (2) in equation (1), we get:

P(a) = 2 $\times$ 2 P(c) = 4 P(c)

$\Rightarrow$ P(a) = 4 P(c) ......(3)

Given, P(a) + P(b) + P(c) = 1

Putting values from equation (2) and (3), we get:

$\Rightarrow 4 P(c) + 2 P(c) + P(c) = 1\\\Rightarrow 7 P(c) = 1\\\\\Rightarrow P(c) = \frac{1}{7}$

Now, putting the value of P(c) in equation (2) and (3), we get:

$P(a) = 4 \times \frac{1}{7} = \frac{4}{7}\\\\ \& P(b) = 2 \times \frac{1}{7} = \frac{2}{7}$

Therefore,

Winning probability of a = P(a) = 4/7

Winning probability of b = P(b) = 2/7

And, Winning probability of c = P(c) = 1/7

Answer 2

Answer:

P(A)= 7 /4  ;P(B)=  7 /2  and P(C)=  7 /1

​

Step-by-step explanation:

Let the probability of winning by C i.e., P(C) be x

∴P(B)=2x and

P(A)=2(2x)=4x

Given that P(A)+P(B)+P(C)=1

4x+2x+x=1

7x=1

x=  7 /1

Hence, P(A)= 7 /4  ;P(B)=  7 /2  and P(C)=  7 /1

​