Page 351 - DECO504_STATISTICAL_METHODS_IN_ECONOMICS_ENGLISH

Statistical Methods in Economics

Notes Let B be the event that the second ball drawn is black. Then the conditional event
(B|A) implies drawing a black ball from the box which contains 4 black and 4 white
balls.

4 1
Thus P (B|A) = = .
8 2
The event implies that both the balls drawn are black.
We are required to find P (AB). We have
P (AB) = P (B|A)· P (A)

51 5
= ⋅ = .
92 18
In the above example, consider that the ball drawn in the first draw is returned to the
box, so that the composition of the box remains the same before each drawing. Here it
is drawing with replacement. Now the conditional event (B|A), that the second ball is
black is not affected by the event that the first draw resulted in a black ball since the
ball drawn was returned. Thus the event B and the event A are independent and the
knowledge of one does not affect the other. Then P (B|A) = P (B), independent of P
(A). We have then

P(AB)
P (B) =
P(A)
Or, P (AB) = P (A) P (B) ... (6)
which gives the multiplication rule for independent events, A and B.
Example 11: What is the probability that in 2 throws of a die, six appears in both the dice ?
Let A and B be the event that 6 appears in the first and the second throws respectively;
these events are independent. Then the event AB implies that six appears in both the
dice.

11 1
P (AB) = P (A) P (B) = ⋅ = .
66 36
11 1 1
Example 12: The probability of getting HHT in tossing 3 coins is thus ⋅ ⋅ = , since the three
22 2 8
events Head in first throw, Head in second and Tail in third are independent.
Example 13: Find the probability that in a family of 2 children (i) both are boys, (ii) both are of the
same sex, assuming that the probability of a child being a boy or a girl is equal
⎛ 1 ⎞
⎜ equal to ⎟
⎝ 2 ⎠
(i) Let A , A be the events respectively that the first and the second child is a boy.
1 2
Then since A , A are independent,
1 2
11 1
P (A A ) = P (A ) P (A ) = ⋅ =
1 2 1 2 22 4
Similarly if B , B are the events respectively that the first and the second child
2
1
is a girl, then B , B are independent and
1 2
11 1
P (B B ) = P (B ) P (B ) = ⋅ =
1 2 1 2 22 4

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