What is Probability?

Probability

If in any random experiment the n outcomes are exhaustive, mutually exclusive and equally likely and m of these are favorable to an event A, then the probability of A is defined by

P(A)= m/n

Let A' be the complementary event to A. The favorable outcomes to A' are n-m. Then, probability of A' is given by 

P(A')= n-m/n

        =1-m/n

        =1-P(A)

P(A) + P(A')= 1 

Example 6.1 : In a family there are 3 male and 2 female members. A family work is to be finished by any 2 of them. Find the probability that

(a) the work will be finished by one male and one female member.

(b) the work will be finished by either two males or by two females.

 

Solution: 

(a)

Let A be the event that the work will be finished by one male and one female members. Since the work is to be finished by the two members, it can be finished in

N= ⁵C₂ = 10 ways

The work can be finished by one male and one female member in

m= ³C₁ X ²C₁= 6 ways

P(A) = m/n = 6/10 = 0.6

 

(b) Let A' be the event that the work will be finished either by two male members or by two female members. Here A' is complementary event, since statement of A' is against the statement of A.

P(A') = 1 – P(A) = 1 – 0.6 = 0.4

Example 6.2 : Two unbiased dice are thrown once. Find the probability that

(i) both dice show same number,

(ii) first dice shows even number,

(iii) both dice show even number,

(iv) sum of the upper faces of the dice is 8 or more,

(v) sum of the upper faces of the dice is above 10,

(vi) sum of the upper faces of the dice is less than 7,

(vii) second die shows number 5 or more.

Solution: The sample space of  the experiment is 

S=

11    21    31    41    51    61

12    22    32    42    52    62

13    23    33    43    53    63

14    24    34    44    54    64

15    25    35    45    55    65

16    26    36    46    56    66

n = 36

(i)  Let A be the event that dice show same number. Favorable cases to A, m=6

P(A) = m/n = 6/36 = 1/6

(ii) Let B be the event that first dice shows even number. Favorable cases to B, m=18

P(A) = m/n = 18/36 = 1/2

(iii) Let C be the event that both dice show even number. Favorable cases to C, m=9

P(C) = m/n = 9/36 = 1/4

(iv) Let D be the event that sum of the upper faces of the dice is 8 or more. Favorable cases to D, m=15

P(D) = m/n = 15/36 = 5/12

(v) Let E be the event that sum of upper faces of dice is above 10. Favorable cases to E, m=3

P(E) = m/n = 3/36 = 1/12

(vi) Let F be the event that sum of upper faces of dice is less than 7. Favorable cases to F, m=15

P(F) = m/n = 15/36 = 5/12

(vii) Let G be the event that second dice shows number 5 or more. Favorable cases to G, m=12

P(G)= m/n = 12/36 = 1/3

Example 6.3: Three biased coins are tossed once. It is known that any coin shows head with probability P(H) = 2/3. [ P(T) = 1/3 ]. Find the probability that

(i) ta least two heads will appear

(ii) at best two heads will appear

(iii) there will be no head

(iv) there will be two heads

Solution:

The sample space of the experiment is

S= {HHH, HHT, HTT, HTH, TTH, THH, THT, TTT}

Given P(H) = 2/3, P(T) = 1/3

(i) Let A be the event that at least two heads will appear. Favorable points to A are 

A= {HHH, HHT, HTH, THH}

P(H) = P(HHH)+P(HHT)+P(HTH)+P(THH)

        =2/3*2/3*2/3 + 2/3*2/3*1/3 + 2/3*1/3*2/3 + 1/3*2/3*2/3

        = 20/27

(ii) Let B be the event that will be two heads. Then B' is the event showing 3 heads.

P(B') = P(HHH) = 2/3*2/3*2/3 = 8/27

P(B) = 1-P(B') = 1 - 8/27 = 19/27

(iii) Let C be the event showing no head. Favorable point to C is 

C = {TTT}

P(C) = P(TTT) = 1/3*1/3*1/3 = 1/27

(iv) Let D be the event showing 2 heads. Favorable cases to D are

D = {HHT, HTH, THH} 

P(D) = P(HHT) + P(HTH) + P(THH)

        = 2/3*2/3*1/3 + 2/3*1/3*2/3 + 1/3*2/3*2/3

        =12/27