The above symbols represent the 52 cards of a deck of playing cards. If a card is selected at random from the deck then each of the 52 signs above is a member of the sample space. So the sample space in selecting a card  from a pack of playing cards at random consists of 52 members/sample points. Now, suppose that I select a card from the deck at random and then see the sign of the card. You cannot see it but I tell you that it is a Jack (J) card. Can you tell me whether or not it is spades? Of course not. A Jack card may be the one of spades, clubs, diamonds, or hearts. Let A = a Jack card is selected and B = a spades card is selected. Statistically speaking, A and B are independent events. In other words, if we know that A occurs, we don’t have any idea whether or not B occurs. Also, if we know that B occurs, we don’t have any idea whether or not A occurs. Stated another way, A has no “predictive power” at all about the occurrence of B. Also, B has no predictive power at all about the occurrence of A.

Compare with the following situation. Let C = a card with red symbol is selected and D = a hearts card is selected. Suppose that I select a card from the deck at random. I can see the sign of the card but you can’t. Then, I tell you that the card has a red symbol and you are asked to guess the symbol, whether it is a spades, clubs, diamonds, or hearts. If you are rational, your answer will not be spades or clubs because they are of black color. Your guess may be diamonds or hearts. Statistically speaking, In this case C and D are not independent events. The information that C occurs or doesn’t occur is “useful” in predicting whether or not it is a hearts card. In other words, information about the occurrence of C has “predictive power” about the occurrence of D.

 

Definition

Let A and B be possible events, i.e. P(A), P(B) > 0. A and B are independent events if the following condition is satisfied: P(A \cap B) = P(A) \cdot P(B). Otherwise, the events are said to be dependent.

 

Let’s prove that in the first case, the events are independent. Note that the number of sample points of S is 52, that is \mid S \mid = 52.

Therefore, \mid A \mid = 4 , P(A) = \frac{4}{52} = \frac{1}{13}.
\mid B \mid = 13 , P(B) = \frac{13}{52} = \frac{1}{4}
\mid A \cap B \mid = 1 , P(A \cap B) = \frac{1}{52}
Note that \frac{1}{52} = \frac{1}{13} \cdot \frac{1}{4}, i.e. P(A \cap B) = P(A) \cdot P(B).
We conclude that A and B are independent.

 

Now, let’s turn to the second case. We’ll prove that the events are dependent. The events can be represented as:

Therefore, \mid C \mid = 26$ , $P(C) = \frac{26}{52} = \frac{1}{2}.
\mid D \mid = 13 , P(D) = \frac{13}{52} = \frac{1}{4}
\mid C \cap D \mid = 13 , P(C \cap D) = \frac{13}{52} = \frac{1}{4}
Note that \frac{1}{4} \neq \frac{1}{2} \cdot \frac{1}{4}, i.e. P(C \cap D) \neq P(C) \cdot P(D).
So, C and D are not independent events. They are dependent.

 

Conditional Probability

If A and B are dependent events, we use the formula P(A \cap B) = P(A) \cdot P(B \mid A) to find the probability of the simultaneous occurrence of A and B. P(B \mid A) is the probability that B occurs if A occurs. Stated another way, P(B \mid A) is the probability that B occurs, assuming A occurs. P(B \mid A) is a conditional probability.

Suggestion:

If we are not sure whether the events are independent or not, it is safer to apply P(A \cap B) = P(A) \cdot P(B \mid A).

 

Sample Problem 1
A box contains 4 red and 6 green balls. Two balls are drawn in succession. Before drawing the second ball, the first is not put back into the box. (It is often called sampling without replacement.) Find the probability that we get a green ball at the first draw and a red at the second.

Answer
Let A = a green ball is selected at the first draw and B = a red ball is selected at the second draw. What we have to find is the probability that (both) A and B occur, i.e. P(A \cap B). The appropriate formula to apply is P(A \cap B) = P(A) \cdot P(B \mid A).
As there are 6 green available before the first drawing, the probability of getting a green ball at the first draw is P(A) = \frac{6}{10} = \frac{3}{5}. To calculate P(B \mid A), we have to assume that A occurs. (So, we assume that we get a green ball at the first drawing.) As a consequence of the assumption, now there are 4 red and 5 green balls in the box just before the second drawing is performed. Now, the probability of getting a red ball is \frac{4}{9}. We denote it as P(B \mid A) = \frac{4}{9}. Applying the formula, we have:
P(A \cap B) = P(A) \cdot P(B \mid A) = \frac{3}{5} \cdot \frac{4}{9} = \frac{4}{15}.
So, the probability of getting a green ball at the first draw and a red at the second is \frac{4}{15}.

 

Sample Problem 2
A box contains 4 red and 6 green balls. Two balls are drawn in succession. Before drawing the second ball, the first is put back into the box. (It is called sampling with replacement.) Find the probability that we get a green ball at the first draw and a red at the second.

Answer
Let A = green ball is selected at the first draw and B = a red ball is selected at the second draw. What we have to find is the probability that (both) A and B occur, i.e. P(A \cap B). The suitable formula to use is P(A \cap B) = P(A) \cdot P(B \mid A). As there are 6 green balls available before the first drawing, the probability of getting a green ball at the first draw is P(A) = \frac{6}{10} = \frac{3}{5}. To calculate P(B \mid A), we have to assume that A occurs. (So, we assume that we get a green ball at the first drawing.) The assumption implies that there are 4 red and 5 green balls in the box now. However, the green ball is put back into the box, so the initial condition is restored. As consequence, just before the second drawing is performed there are still 4 red and 6 green balls. Therefore, the probability of getting a red ball (at the second draw) is \frac{4}{10} = \frac{2}{5}. We denote it as P(B \mid A) = \frac{2}{5}. Applying the formula, we have:
P(A \cap B) = P(A) \cdot P(B \mid A) = \frac{3}{5} \cdot \frac{2}{5} = \frac{6}{25}. So, the probability of getting a green ball at the first draw and a red at the second is \frac{6}{25}.

 

Sample Problem 2 can be solved more briefly if we know that the events are independent. If the events are independent, P(A \cap B) = P(A) \cdot P(B).
P(A) = \frac{6}{10} = \frac{3}{5}
P(B) = \frac{4}{10} = \frac{2}{5}
P(A \cap B) = P(A) \cdot P(B) = \frac{3}{5} \cdot \frac{2}{5} = \frac{6}{25}

Note:
If A and B are independent events then A has no influence on the occurrence of B, therefore P(B \mid A) = P(B). Consequently,
P(A \cap B) = P(A) \cdot P(B \mid A)
P(A \cap B) = P(A) \cdot P(B) ( because P(B \mid A) = P(B) )
That’s why the different methods above have resulted in the same conclusion, i.e. P(A \cap B) = \frac{6}{25}.

 

Sample Problem 3
In the experiment of rolling two dice once, what is the probability that we get a sum of 10 if both dice show an equal number of spots on the sides facing up.

Answer
Let A = the dice show an equal number of spots and B = a sum of 10 appears. What we have to find is the probability of getting a sum of 10 if the dice show an equal number of spots, i.e. P(B \mid A). Note that P(A \cap B) = P(A) \cdot P(B \mid A) is equivalent to P(B \mid A) = \frac{P(A \cap B)}{P(A)}. The sample space is S = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} and \mid S \mid = 36. Furthermore:
A \cap B = \{ (5,5) \} , \mid A \cap B \mid = 1 , P(A \cap B) = \frac{1}{36}
A = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}, \mid A \mid = 6 , P(A) = \frac{6}{36} = \frac{1}{6}
P(B \mid A) = \frac{P(A \cap B)}{P(A)} = \frac{1/36}{1/6} = \frac{1}{6}.

 

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