Multiplication Law of Probability
If A and B are any two independent or dependent events
then the occurrence of events A and B (i.e. A∩B) in two successive trials is given
by the following laws of multiplication of probabilities.
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Multiplication law of probability for independent events:
Two or more events are said to be independent if the
occurrence of one of the events does not affect the occurrence of other events.
For example, in the random experiment of tossing a coin twice or more, the
occurrence of any one event in the first trial does not affect the occurrence
of any event in the second trial.
If A and B are any two independent events, then the probability
of the occurrence of A and B (i.e. A∩B) in
two successive trials is equal to the product of the individual probabilities
of the occurrence of A and B in two trials.
i.e. P(A and B) = P(A) × P(B)
i.e. P(A∩B)
= P(A) × P(B)
Further more, if A, B and C are three independent events,
then,
P(A
and B and C) = P(A∩B∩C)
= P(A) × P(B) × P(C)
and
so on.
Multiplication law of probability for dependent events:
Two or more events are said to be dependent if the
occurrence of one of the events affects the occurrence of the other events. For
example, while drawing a ball in two successive trials from a bag containing 2
red and 3 green without a replacement, getting any one coloured ball in the
first trial affects to draw another ball in the second trial. So these are the
dependent events.
If A and B are any two dependent events, then the
probability of the occurrence of A and B in two successive trials is given by,
P(A and
B) = P(A∩B)
= P(A) × P(B/A)
Further more, if A, B and C are three independent events,
then,
P(A and B and C) = P(A∩B∩C)
= P(A) × P(B/A) × P(C/A and B)
and so on.
Note:

P(B/A) is the probability of event B given
that A has occurred.

P(C/A and B) is the probability of C given
that A and B has occurred.
Worked out examples on multiplication law of probabilities:
Example 1: Two cards are drawn one after another with replacement
from a well shuffled pack of 52 cards. Find the probability that both cards are
king.
Solution:
There are 4
kings in a pack of 52 playing cards. Let, K denotes the event of king. ∴ n(K) = 4, n(S) = 52
Since the
card which is drawn at first is replaced to draw another card, it is an
independent event.
Now, by
using multiplication law of probability for independent events,
P(K and K) = P(K∩K) = P(K) × P(K)
= n(K)/n(S) × n(K)/n(S)
= 4/52 × 4/52
= 1/13 × 1/13
= 1/169
So, the required probability is 1/169.
Example 2: A bag contains 5 red balls and 3 blue balls. A ball is
drawn at random and replaced. After that another ball is drawn. Find the
probability that: (i) both balls are
blue (ii) none of them are blue.
Solution:
Let, R
denotes red and B denotes blue balls, ∴ n(R) = 5, n(B)= 3
The total
number of possible outcomes, n(S) = total number of balls
= 5 + 3
= 8
Since, the
ball which is drawn at first is replaced to draw another ball, it is an
independent event.
Now, by
using multiplication law of probability for independent events,
(i)
P(B and B) = P(B∩B) = P(B) × P(B)
= n(B)/n(S) ×
n(B)/n(S)
= 3/8 × 3/8
= 9/64
So, the probability that both of them are blue is 9/64.
(ii)
P(B^{c} and B^{c})
= P(B^{c}∩B^{c}) = P(B^{c})
× P(B^{c})
= [1 – P(B)] × [1 – P(B)]
= [1 – 3/8] × [1 – 3/8]
= 5/8 × 5/8
= 25/64
So, the probability that none of them are blue is 25/64.
Example 3: A box contains 4 red, 3 blue and 5 white balls. A ball is
drawn at random and it is replaced, then another ball is drawn. Find the
probability that:
(i)
the first is white and the
second is blue,
(ii)
the first is blue and the
second is red,
(iii)
both of them are red,
(iv)
both of them are of the same
colour,
(v)
both of them are not blue.
Solution:
Let, R
denotes red, B denotes blue and W denotes white balls, ∴ n(R) = 4, n(B) = 3,
n(W) = 5
The total
number of possible outcomes, n(S) = total number of balls
= 4 + 3 + 5
= 12
Since, the
ball which is drawn at first is replaced to draw another ball, it is an
independent event.
Now, by
using multiplication law of probability for independent events,
(i)
P(W and B) = P(W∩B) = P(W) × P(B)
= n(W)/n(S) ×
n(B)/n(S)
= 5/12 × 3/12
= 5/48
So, the probability that the first is white and the second is blue is 5/48.
(ii)
P(B and R) = P(B∩R) = P(B) × P(R)
= n(B)/n(S) × n(R)/n(S)
= 3/12 × 4/12
= 1/12
So, the probability that the first is blue and the second is red
is 1/12.
(iii)
P(R and R) = P(R∩R) = P(R) × P(R)
= n(R)/n(S) ×
n(R)/n(S)
= 4/12 × 4/12
= 1/9
So, the probability that both of them are red is 1/9.
(iv)
P(RR or BB or WW) =
[P(R) × P(R)] + [P(B) × P(R)] + [P(W) × P(W)]
= n(R)/n(S) × n(R)/n(S) + n(B)/n(S) × n(B)/n(S) + n(W)/n(S) ×
n(W)/n(S)
= 4/12 × 4/12 + 3/12 × 3/12 + 5/12 × 5/12
= 1/9 + 1/16 + 25/144
= 50/144
= 25/72
So, the probability that both of them are of same colour is 25/72.
(v)
P(B^{c} and B^{c})
= P(B^{c}∩B^{c}) = P(B^{c})
× P(B^{c})
= [1 – P(B)] × [1 – P(B)]
= [1 – 3/12] × [1 – 3/12]
= [1 – 1/4] × [1 – 1/4]
= 3/4 × 3/4
= 9/16
So, the probability that
both of them are not blue is 9/16.
Example 4: If two unbiased coins are tossed simultaneously, find the
probability of getting (i) two heads (ii) at least one head.
Solution:
Here, for each
coin, n(S) = 2, n(H) = 1 and n(T) = 1
Since, the
result obtained for both the coins are independent, it is the case of
independent event. So, by using the multiplication law for independent events,
(i)
P(H and H) = P(H∩H) = P(H) × P(H)
= n(H)/n(S) × n(H)/n(H)
= ½ × ½
= ¼
So, the probability of two heads is ¼.
(ii)
P(HH or HT or TH) = P(H)
× P(H) + P(H) × P(T) + P(T) × P(H)
= ½ × ½ + ½ × ½ + ½ ×
½
= ¼ + ¼ + ¼
= ¾
So, the probability of at least one head is ¾.
Example 5: Find the probability of getting 3 on the dice and head on
the coin when a dice is rolled and a coin is tossed simultaneously.
Solution:
Here, for a
dice, n(S_{1}) = 6 , n(3) = 1 and for a coin, n(S_{2}) = 2,
n(H) = 1
It is the
case of independent event. So, by using the multiplication law for independent
events,
P(3 and H) = P(3) × P(H)
= n(3)/n(S_{1}) × n(H)/n(S_{2})
= 1/6 × 1/2
= 1/12
So, the
required probability is 1/12.
Example 6: The probability of solving a mathematical problem by two
students A and B are 1/3 and 1/4 respectively. If the problem is given to the
both students, find the probability of solving.
Solution:
Here, the
problem may be solved by both the students. So, it is the case of nonmutually
exclusive event.
∴ P(A or B) = P(A∪B) = P(A) + P(B) – P(A∩B)
Also, the
problem solving event is independent.
∴ P(A and B) = P(A∩B) = P(A) ×
P(B)
= 1/3 × 1/4
= 1/12
Now,
P(A∪B) = P(A) + P(B) – P(A∩B)
=
1/3 + 1/4 – 1/12
=
6/12
=
1/2
So, the
required probability is 1/12.
Example 7: A box contains 5 red and 3 white balls. Two balls are
drawn randomly one by one without replacement. Find the probability of:
(i)
both the balls are red
(ii)
both the balls are white
(iii)
first red and second ball is
white
(iv)
first white and second ball
is red
Solution:
Let, R
denotes red and W denotes white balls. Total no. of balls at first = 5 + 3 = 8.
Since, the ball which is drawn at first is not replaced to draw another ball,
it is an dependent event.
Now, by
using multiplication law of probability for dependent events,
(i)
P(R and R) = P(R∩R) = P(R) × P(R/R)
= 5/8 × 4/7
= 5/14
So, the probability of both the balls are red is 5/14.
(ii)
P(W and W) = P(W∩W) = P(W) × P(W/W)
= 3/8 × 2/7
= 3/28
So, the probability that of both the balls are white is 3/28.
(iii)
P(R and W) = P(R∩W) = P(R) × P(W/R)
= 5/8 × 3/7
= 15/56
So, the probability of first red and second ball is white is 15/56.
(iv)
P(W and R) = P(W∩R) = P(W) × P(R/W)
= 3/8 × 5/7
= 15/56
So, the probability of
first white and second ball is red is 15/56.
Example 8: There are 3 black and 5 white balls in a box. Two balls
are drawn randomly one by one without replacement. Find the probability of both
the balls are of same colour.
Solution:
Let, B
denotes black and W denotes white balls. Total no. of balls at first = 3 + 5 =
8. Since, the ball which is drawn at first is not replaced to draw another
ball, it is an dependent event.
Now, by
using multiplication law of probability for dependent events,
P(B and B) = P(B∩B) = P(B) ×
P(B/B)
= 3/8 × 2/7
= 3/28
P(W and W) = P(W∩W) = P(W) ×
P(W/W)
= 5/8 × 4/7
= 5/14
Again, the
probability of both the balls are of same colour is,
P(BB or WW) = P(BB) + P(WW)
= 3/28 + 5/14
= (3 + 10)/28
= 13/28
So, the probability
of both the balls are of same colour is 13/28.
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