Multiplication Law of Probability

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)

Furthermore, 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 anyone coloured ball in the first trial affects the drawing of 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(S₁) = 6 , n(3) = 1 and for a coin, n(S₂) = 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(S₁) × n(H)/n(S₂)

                        = 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 non-mutually 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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