In a pet store, there are 8 puppies, 10 kittens and 6 parakeets. If a pet is chosen at random, what is the probability of choosing a puppy or a kitten?1234144

Answer :

FORMULAS

If A and B are any events then:

[tex]P\mleft(AorB\mright)=P\mleft(A\mright)+P\mleft(B\mright)-P\mleft(AandB\mright)[/tex]

If A and B are mutually exclusive events then P(A and B) = 0, so then:

[tex]P\mleft(A\text{ }or\text{ }B\mright)=P\mleft(A\mright)+P\mleft(B\mright)[/tex]

The probability of an event happening is given to be:

[tex]P=\frac{\text{Number of required events}}{Number\text{ of total events}}[/tex]

SOLUTION STEPS

The total number of pets in the store is calculated to be:

[tex]Total=8+10+6=24[/tex]

The probability of picking a puppy is:

[tex]P(p)=\frac{8}{24}[/tex]

The probability of picking a kitten is:

[tex]P(k)=\frac{10}{24}[/tex]

Therefore, the probability of choosing a puppy or a kitten is gotten to be:

[tex]P(p\text{ }or\text{ }k)=\frac{8}{24}+\frac{10}{24}=\frac{18}{24}=\frac{3}{4}[/tex]

ANSWER

The probability is 3/4 or 0.75.

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