# Thread: random selection, probability.

1. ## random selection, probability.

55% of all people use cash and 38% of all use an ATM card. research has shown that those paying with cash, 75% use discounts, and those using an ATM card, 35% use discounts, and those using a credit card, just 10% use discounts.
Suppose a random person is randomly selected.

a.) What if the person does not use discounts, what is the probability that this person paid with an ATM card?

...I'm having trouble setting this one up.

....however I do know the probability of someone who does not use coupons.

(.55*.25) + (.07*.90) + (.38*.65)
=.4475

I tried to use this to help me solve my question above, but still want help.

2. ## Re: random selection, probability.

Originally Posted by rcmango
55% of all people use cash and 38% of all use an ATM card. research has shown that those paying with cash, 75% use discounts, and those using an ATM card, 35% use discounts, and those using a credit card, just 10% use discounts.
Suppose a random person is randomly selected.

a.) What if the person does not use discounts, what is the probability that this person paid with an ATM card?

...I'm having trouble setting this one up.

....however I do know the probability of someone who does not use coupons.

(.55*.25) + (.07*.90) + (.38*.65)
=.4475

I tried to use this to help me solve my question above, but still want help.
Let A be the event "Paid with ATM card".
Let B be the event "Does not use discounts."
Calculate the conditional probability Pr(A|B).

3. ## Re: random selection, probability.

Hello, rcmango!

55% of all people use cash and 38% of all use an ATM card.
. . and other 7% use credit cards.

Research has shown that those paying with cash, 75% use discounts,
and those using an ATM card, 35% use discounts,
and those using a credit card, just 10% use discounts.

Suppose a random person is randomly selected.

a) If the person selected does not use discounts,
. . what is the probability that this person paid with an ATM card?

... I'm having trouble setting this one up.

.... However, I do know the probability of someone who does not use coupons.

. . $\displaystyle (0.55\cdot0.25) + (0.07\cdot0.90) + (0.38\cdot0.65) \,=\,0.4475$ . Good!

I tried to use this to help me solve my question above, but still want help.

You already found these probabilities:

. . $\displaystyle \begin{array}{cccccc}P(\text{Cash }\wedge \sim\text{discount}) &=& (0.55)(0.25) &=& 0.1375 \\ P(\text{ATM }\wedge \sim\text{discount}) &=& (0.38)(0.65) &=& 0.2470 & [1] \\ P(\text{Card }\wedge \sim\text{discount}) &=& (0.07)(0.10) &=& 0.0630 \\ &&&& ---- \\ && P(\sim\text{disc}) &=& 0.4475 & [2] \end{array}$

As mr fantastic pointed out, use the Conditional Probability Formula.

. . $\displaystyle P(\text{ATM}\,|\,\sim\text{discount}) \;=\;\frac{P(\text{ATM } \wedge \sim\text{discount})}{P(\sim\text{discount})}$

Just divide [1] by [2].