There is a 0.23 prob. that typical customer buys gasoline...

[ravifour]

There is a 0.23 probability that a typical convenience store customer buys gasoline. The probability that a customer buys groceries is 0.76 and the conditional probability of buying groceries given that the customer buys gasoline is 0.85.

a. Find the probability that a typical customer buys both gasoline and groceries.

My Ans:
P(Gas) = .23
P(Gro) = .76
P(Gro|Gas) = .85 = P(Gro n Gas)/P(Gas)

P(Gro n Gas) = P(Gro|Gas) * P(Gas) = .85 *.23 = .1955

b. Find the probability that a typical customer buys gasoline or groceries. (5)

My Ans:
P(Gro u Gas) = P(Gro) + PGas) - P(Gro n Gas) = .23 + .76 -.1955 = .7945

Please correct me if I am wrong.

 

[soroban]

Re: another question

Hello, ravifour!

There is a 0.23 probability that a typical convenience store customer buys gasoline.
The probability that a customer buys groceries is 0.76
and the conditional probability of buying groceries given that the customer buys gasoline is 0.85

a. Find the probability that a typical customer buys both gasoline and groceries.

My answer:\(\displaystyle \(\text{Gas}) \,= \,0.23,\;P(\text{Gro})\, =\, 0.76\)

\(\displaystyle P(\text{Gro}\,|\,\text{Gas}) \:= \:\frac{P(\text{Gro}\,\cap\,\text{Gas})}{P(\text{Gas})}\;\;\Rightarrow\;\;0.85\:=\:\frac{P(\text{Gro}\,\cap\,\text{Gas})}{0.23}\)

\(\displaystyle P(\text{Gro}\,\cap\,\text{Gas}) \:= \(\text{Gro}\,|\,\text{Gas})\,\cdot\,P(\text{Gas}) \:= \0.85)(0.23) \:= \:0.1955\)


b. Find the probability that a typical customer buys gasoline or groceries. (5)

My answer:

\(\displaystyle \(\text{Gro}\,\cup\,\text{Gas}) \:= \(\text{Gro})\,+\,P(\text{Gas})\,-\,P(\text{Gro}\,\cap\,\text{ Gas})\: =\:0.23\,+\,0.76\,-\,0.1955 \:=\:0.7945\)

As we say in my college faculty (or was it the Marine Corps?): Ya done good!

 

[ravifour]

Soroban Thank you

Soroban Thank you, for clarifying.