Question

In a certain race there are three girls X,Y,Z.
The winning probability of X is twice than Y and the winning probability of Y is twice than Z. If P(X)+P(Y)+P(Z) =1,then find the winning probability of each girl.​

Answer 1

Answer:

let probability of Z be'a' i.e. P(Z)=a

probability of Y is twice than Z i.e.P( Y)=2a

probability of X is twice than Y i.e. P(X)=2Y=2(2a)=4a

P(X)+P(Y)+P(Z)=1

4a+2a+a=1

7a=1

a=1/7=P(Z)

P(Y)=2(1/7)=2/7

P(X)=4(1/7)=4/7

Answer 2

Answer:

x=4/7

y=2/7

z= 1/7

Step-by-step explanation:

x=winning probability is =x

y=x/2

z=x/4

s.t.q ( x/2) + (X/4)+x =1

2x+4x+x =4

7x =4

x=4/7

y=2/7

z= 1/7

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