1. ## Dependent events Probability

Two problems:
1. Hannah picks two cards from a bowl with 3 orange, 4 purple, and 3 white cards. She does not return them to the bowl after picking them. Find P(the 1st card is not orange and the 2nd card is not white). The first part is simple 7/10 dependent part is not so clear. There are two ways this problem can happen. Choose a purple or a white to satisfy first condition. so you get P(purple)*P(not white) + P(white)*P(not white) leading to the final answer:
7/10*(4/10*6/9+3/10*7/10)

Is that right?

2. What is the probability that you draw a heart from deck of 52 cards don't replace it and then draw an ACE?

13/52* 4/51

Is that right or do I need to account somehow for possibility of an ace of hearts being the first card?

2. Hello, LArchambeau3!

In both problems, the probability of the second event depends on the first event.

1. Hannah picks two cards from a bowl with 3 orange, 4 purple, and 3 white cards.
She does not return them to the bowl after picking them.

Find P(the 1st card is not orange and the 2nd card is not white).
There are two cases to consider:

. . [1] The first card is white and the second is not white.
. . [2] The first card is purple and the second is not white.

[1] First is White, second is not-White.

. . $\displaystyle P(\text{1st White}) \:=\:\frac{3}{10}$

. . Now there are: 3 Orange, 4 Purple, and 2 White in the bowl.

. . $\displaystyle P(\text{2nd not-White}) \:=\:\frac{7}{9}$

. . Hence: .$\displaystyle P(\text{1st White} \wedge\text{2nd not-White}) \:=\:\frac{3}{10}\cdot\frac{7}{9} \:=\:\frac{21}{90}$

[2] First is Purple, second is not-White.

. . $\displaystyle P(\text{1st Purple}) \:=\:\frac{4}{10}$

. . Now there are: 3 Orange, 3 Purple, 3 White in the bowl.

. . $\displaystyle P(\text{2nd not-White}) \:=\:\frac{6}{9}$

. . Hence: .$\displaystyle P(\text{1st Purple}\wedge\text{2nd not-White}) \:=\:\frac{4}{10}\cdot\frac{6}{9} \:=\:\frac{24}{90}$

Therefore: .$\displaystyle P(\text{1st not-Orange} \wedge \text{2nd not-White}) \;=\;\frac{21}{90} + \frac{24}{90} \:=\:\frac{45}{90} \:=\:\frac{1}{2}$

2. What is the probability that you draw a Heart from a deck of 52 cards,
don't replace it, and then draw an Ace?
There are two cases to consider:

. . [1] The first card is the Ace of Hearts.
. . [2] The first card is some other Heart.

[1] The first card is the Ace of Hearts, the second is another Ace.

. . $\displaystyle P(\text{1st A}\heartsuit) \:=\:\frac{1}{52}$

. . There are 51 card left, 3 of them are Aces.

. . $\displaystyle P(\text{2nd Ace}) \:=\:\frac{3}{51}$

. . Hence: .$\displaystyle P(\text{1st A}\heartsuit \wedge \text{2nd Ace}) \:=\:\frac{1}{52}\cdot\frac{3}{51} \:=\:\frac{3}{2652}$

[2] The first card is some other heart, the second is an Ace.

. . $\displaystyle P(\text{1st other Heart}) \:=\:\frac{12}{52}$

. . There are 51 cards left, 4 of them are Aces.

. . $\displaystyle P(\text{2nd Ace}) \:=\:\frac{4}{51}$

. . Hence: .$\displaystyle P(\text{1st other Heart} \wedge\text{2nd Ace}) \;=\;\frac{12}{52}\cdot\frac{4}{51} \:=\:\frac{48}{2652}$

Therefore: .$\displaystyle P(\text{1st Heart} \wedge \text{2nd Ace}) \;=\;\frac{3}{2652} + \frac{48}{2652} \;=\;\frac{51}{2652} \;=\;\frac{1}{52}$