In a race where 12 cars are running- the chance that car A will win is 1-8- that B will win is 1-10 and that C will win is 1-6- Find the chance that one of them will win-

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Dependent Events

In a race whe...

Question

In a race where 12 cars are running, the chance that car A will win is $\frac{1}{8}$ , that B will win is $\frac{1}{10}$ and that C will win is $\frac{1}{6}$ . Find the chance that one of them will win.

1/6

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1/10

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1/8

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47/120

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None of these

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Solution

The correct option is D
47/120

Required probability = P(X)+P(Y)+P(Z) (all the events are mutually exclusive).
= $\frac{1}{6}$ + $\frac{1}{10}$ + $\frac{1}{8}$ = $\frac{47}{120}$

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In a race where 12 cars are running, the chance that car A will win is 18, that B will win is 110 and that C will win is 16. Find the chance that one of them will win.

The correct option is D 47/120 Required probability = P(X)+P(Y)+P(Z) (all the events are mutually exclusive). = 16 + 110 + 18 = 47120

In a race where 12 cars are running, the chance that car A will win is $\frac{1}{8}$ , that B will win is $\frac{1}{10}$ and that C will win is $\frac{1}{6}$ . Find the chance that one of them will win.

The correct option is D
47/120

Required probability = P(X)+P(Y)+P(Z) (all the events are mutually exclusive).
= $\frac{1}{6}$ + $\frac{1}{10}$ + $\frac{1}{8}$ = $\frac{47}{120}$