Question 1199784: Suppose that you have 7 green cards and 5 yellow cards. The cards are well shuffled. You randomly draw two cards without replacement.
G1 = the first card drawn is green
G2 = the second card drawn is green
a. P(G1 and G2) =
b. P(At least 1 green) =
c. P(G2|G1) =
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!
Part (a)
G1 = the first card drawn is green
G2 = the second card drawn is green
P(G1) = probability of the first card being green
P(G1) = (7 green)/(7+5 total)
P(G1) = (7 green)/(12 total)
P(G1) = 7/12
The key phrasing "without replacement" means that the 2nd selection probability will depend on the 1st selection.
Whatever happens on the 1st draw will change the probability of the 2nd draw.
In short, the events are dependent.
Refer to conditional probability.
P(G2 given G1) = probability event G2 happens, given G1 happened
P(G2 given G1) = probability getting a 2nd green, given 1st is green
P(G2 given G1) = (number of green left)/(number total left)
P(G2 given G1) = (7-1 green left)/(12-1 total left)
P(G2 given G1) = (6 green left)/(11 total left)
P(G2 given G1) = 6/11
P(G1 and G2) = P(G1)*P(G2 given G1)
P(G1 and G2) = (7/12)*(6/11)
P(G1 and G2) = (7*6)/(12*11)
P(G1 and G2) = (7*6)/(6*2*11)
P(G1 and G2) = (7)/(2*11)
P(G1 and G2) = 7/22
Side note: If G1 and G2 were independent, then P(G2 given G1) = P(G2) would be the case. However, G1 and G2 are dependent.
Answer: 7/22
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Part (b)
Define two events A and B
A = "at least one green" aka "one or two green"
B = "No greens" aka "both cards are yellow"
Events A and B are the only two options.
Their probabilities add to 1.
P(A)+P(B) = 1
P(A) = 1-P(B)
Let
Y1 = the first card is yellow
Y2 = the second card is yellow
P(Y1) = (number of yellow)/(number total)
P(Y1) = (5 yellow)/(12 total)
P(Y1) = 5/12
P(Y2 given Y1) = (number of yellow leftover)/(total leftover)
P(Y2 given Y1) = (5-1 yellow leftover)/(12-1 leftover)
P(Y2 given Y1) = (4 yellow leftover)/(11 leftover)
P(Y2 given Y1) = 4/11
P(B) = probability both cards are yellow
P(B) = P(Y1 and Y2)
P(Y1 and Y2) = P(Y1)*P(Y2 given Y1)
P(Y1 and Y2) = (5/12)*(4/11)
P(Y1 and Y2) = (5*4)/(12*11)
P(Y1 and Y2) = (5*4)/(4*3*11)
P(Y1 and Y2) = (5)/(3*11)
P(Y1 and Y2) = 5/33
P(B) = 5/33
P(A) = 1-P(B)
P(A) = 1-(5/33)
P(A) = (33/33)-(5/33)
P(A) = (33-5)/33
P(A) = 28/33
Answer: 28/33
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Part (c)
This was computed earlier in part (a)
The vertical bar means "given"
P(G2 | G1) = P(G2 given G1)
I prefer to use the word "given" rather than the vertical bar, since the vertical bar may be mixed up with the numeric digit one. Or it might be mistaken for an uppercase i, or lowercase L.
Answer: 6/11
Here is a somewhat similar question, except with different numbers, and the cards are selected with replacement.
https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.1199668.html
Let me know if you have further questions.
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