The conditional probability of event E given that event F has occurred, is given by
P(E|F) = P(E∩F)/P(F), P(F) ≠0
⇒ P(E∩F) = P(E|F).P(F) ...(1)
The conditional probability of event F given that event E has occurred, is given by
P(E|F) = P(F∩E)/P(E), P(E) ≠0
⇒ P(F|E) = P(E∩F)/P(E) (∵E ∩ F = F ∩ E)
⇒ P(E∩F) = P(F|E).P(E) ...(2)
From (1) and (2), we get
P(E∩F) = P(E|F).P(F) = P(F|E).P(E)
This is called the multiplication rule of probability.
Ex:
Consider a pack of cards. Two cards are drawn one after the other, without replacing the first card back into the pack.What is the probability of the event, "drawing a king first and then a queen?
Sol:
Let K and Q be the events of drawing a king and a queen card, respectively.
The probability of drawing a king first and then a queen = P(K ∩ Q)
Total number of cards = 52
Number of kings = 4
∴ P(K) = 4/52
Numbers of cards present in the pack = 52 - 1 = 51 (∵ 1 card is already drawn.)
Number of queens = 4
The conditional probability of Q given that K has occurred: P(Q|K) = 4/51
From the multiplication rule of probability,
P(K∩Q) = P(Q|K).P(K)
P(K∩Q) = 4/51 x 4/52
= 4/51 x 1/13 = 4/663
∴ P(K∩Q) = 4/663
The probability of drawing a king first and then a queen is 4/663.
Multiplication rule for three events E, F and G:
P(E ∩ F ∩ G) = P(E).P(F|E).P(G|(E ∩ F)) = P(E).P(F|E).P(G|EF)
Ex:
An urn contains 12 red balls and 8 blue balls.Suppose we take two red balls and then a blue ball one after another without replacement. What would be the probability of the event of taking out two red balls and a blue ball?
Sol:
Let R: Red ball is drawn
B: Blue ball is drawn
Total number of balls available in the urn = 12 red balls + 8 blue balls = 20
Probability of taking one red ball: P(R) = 12/20 = 3/5
Number of red balls available = 12 - 1 = 11
Number of balls available (after a red ball is drawn)= 20 - 1 = 19
Probability of taking second red ball given that one red ball has already been taken: P(R|R) = 11/19
Number of balls available (after two red balls are drawn) = 19 - 1 = 18
Probability of taking a blue ball given that two red balls have already been taken: P(B|RR) = 8/18 = 4/9
P(RRB) = P(R).P(R|R).P(B|RR)
= 3/5 x 11/19 x 4/9
= 1/5 x 11/19 x 4/3 = 44/285
∴ P(RRB) = 44/285
The probability of taking two successive red balls and a blue ball is 44/285.
Summary
The conditional probability of event E given that event F has occurred, is given by
P(E|F) = P(E∩F)/P(F), P(F) ≠0
⇒ P(E∩F) = P(E|F).P(F) ...(1)
The conditional probability of event F given that event E has occurred, is given by
P(E|F) = P(F∩E)/P(E), P(E) ≠0
⇒ P(F|E) = P(E∩F)/P(E) (∵E ∩ F = F ∩ E)
⇒ P(E∩F) = P(F|E).P(E) ...(2)
From (1) and (2), we get
P(E∩F) = P(E|F).P(F) = P(F|E).P(E)
This is called the multiplication rule of probability.
Ex:
Consider a pack of cards. Two cards are drawn one after the other, without replacing the first card back into the pack.What is the probability of the event, "drawing a king first and then a queen?
Sol:
Let K and Q be the events of drawing a king and a queen card, respectively.
The probability of drawing a king first and then a queen = P(K ∩ Q)
Total number of cards = 52
Number of kings = 4
∴ P(K) = 4/52
Numbers of cards present in the pack = 52 - 1 = 51 (∵ 1 card is already drawn.)
Number of queens = 4
The conditional probability of Q given that K has occurred: P(Q|K) = 4/51
From the multiplication rule of probability,
P(K∩Q) = P(Q|K).P(K)
P(K∩Q) = 4/51 x 4/52
= 4/51 x 1/13 = 4/663
∴ P(K∩Q) = 4/663
The probability of drawing a king first and then a queen is 4/663.
Multiplication rule for three events E, F and G:
P(E ∩ F ∩ G) = P(E).P(F|E).P(G|(E ∩ F)) = P(E).P(F|E).P(G|EF)
Ex:
An urn contains 12 red balls and 8 blue balls.Suppose we take two red balls and then a blue ball one after another without replacement. What would be the probability of the event of taking out two red balls and a blue ball?
Sol:
Let R: Red ball is drawn
B: Blue ball is drawn
Total number of balls available in the urn = 12 red balls + 8 blue balls = 20
Probability of taking one red ball: P(R) = 12/20 = 3/5
Number of red balls available = 12 - 1 = 11
Number of balls available (after a red ball is drawn)= 20 - 1 = 19
Probability of taking second red ball given that one red ball has already been taken: P(R|R) = 11/19
Number of balls available (after two red balls are drawn) = 19 - 1 = 18
Probability of taking a blue ball given that two red balls have already been taken: P(B|RR) = 8/18 = 4/9
P(RRB) = P(R).P(R|R).P(B|RR)
= 3/5 x 11/19 x 4/9
= 1/5 x 11/19 x 4/3 = 44/285
∴ P(RRB) = 44/285
The probability of taking two successive red balls and a blue ball is 44/285.