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On a field trip, there is a 55% chance of kids having pizza for lunch, a 20% chance of kids having tacos for lunch and a 11% chance of kids having pizza and tacos together for lunch. Are the two events "kids eating pizza" and "kids eating tacos" independent? Justify your answer.

2 Answers

4 votes

Answer:

The two events "kids eating pizza" and "kids eating tacos" are independent.

Explanation:

Solution:-

- Denote the following events:

Event ( P ) : kids having pizza for lunch

Event ( T ) : kids having tacos for lunch

- We will interpret each and every statement given in terms of probability of defined events:

- There is a 55% chance of kids having pizza for lunch. Tells us the likely hood of kids having pizza for lunch - Event ( P ) ;

p ( P ) = 0.55

- There is a 20% chance of kids having Tacos for lunch. Tells us the likely hood of kids having Tacos for lunch - Event ( T ) ;

p ( T ) = 0.20

- A 11% chance of kids having pizza and tacos together for lunch. Tells us the likely-hood of two events occurring simultaneously:

p ( T & P ) = 0.11

- We have to investigate whether two defined events ( T ) and ( P ) are independent or not. The condition for independent events is given as:

p ( A & B ) = p ( A ) * p ( B )

- So for the data given to us:

p ( T & P ) = p ( P ) * p ( T )

p ( P ) * p ( T ) = 0.2*0.55 = 0.11

p ( T & P ) = 0.11

Hence,

- The two events Event ( P ) and Event ( T ) are independent events.

User Jean Barmash
by
5.5k points
4 votes

Answer:

Event A is having pizza for lunch and B having tacos for lunch. Since
P(A \cap B) = P(A)P(B), these two events are independent.

Explanation:

Two events, A and B are independent, if:


P(A \cap B) = P(A)P(B)

In this problem:

Event A: having pizza for lunch.

Event B: having tacos for lunch.

55% chance of kids having pizza for lunch

This means that
P(A) = 0.55

20% chance of kids having tacos for lunch

This means that
P(B) = 0.2

11% chance of kids having pizza and tacos together for lunch.

This means that
P(A \cap B) = 0.11

So


P(A \cap B) = P(A)P(B)


0.11 = 0.55*0.2


0.11 = 0.11

Event A is having pizza for lunch and B having tacos for lunch. Since
P(A \cap B) = P(A)P(B), these two events are independent.

User Ikaros
by
5.1k points