2. Meaning of certain terms used in probability
1. Prob. of ATLEAST ONE = Total prob. – Prob. of getting NONE
e.g. Prob of getting at least one head in a tossing of 2 coins
Sample space S = {HH,HT,TH,TT}
At least one head = {HH, HT,TH}
So prob of at least one head = ¾
But total prob = 1, prob of no head = ¼
Total prob – prob of none = 1 – ¼ = 3/4
3. Meaning of certain terms used in probability
PROB. OF ATMOST N items = P(x≤N) = TOTAL PROB – P(x>N)
= P(none) + P(1) + P(2) + P(3) + … + P(N)
= 1 – Prob(x>N)
E.g. Prob of getting at most 5 in a die rolling experiment
Sample space S = {1,2,3,4,5,6}
The event A = {1,2,3,4,5} and so required Prob = n(A)/n(S) = 5/6
Total prob = 1, Prob(x>5) = Prob(x=6) = 1/6
1 - Prob(x>5) = 1 – 1/6 = 5/6
i.e., P(x≤N) = 1 – Prob(x>N)
4. DEPENDENT EVENTS
Two events are said to be dependent, if the happening of one of them
affects the happening of the other. In the case of dependent events the
Chance or prob. one event depends on the happening of other.
E.g. Drawing two cards one by one from a pack of cards without replacement
Here the prob of drawing first card is 1/52.
Since the card is not replaced, the deck of cards contains only 51 cards.
So the prob. of drawing 2nd card is 1/51
5. ASSIGNMENT -
Q1. A bundle of 100 tickets numbered serially from 1 to 100 is well shuffled a ticket
is drawn at random. What is the probability that
a) It is an even number?
b) It is a multiple of 10?
c) It is a perfect square?
d) It is a multiple of 12 or a perfect cube?
e) It is a square or a cube?
f) It is an odd number greater than 75?
h) It is a prime number less than 25?
6. ASSIGNMENT
Q2. A bag contains 5 white, 4 red and 3 black balls. 3 balls are drawn from
the bag at random. Find the probability that
a) there are 2 white and a black balls
b) There is no white ball
c) balls have all the 3 colours
d) none of them are red or black
e) Exactly one white
g) at least one red
H) balls are same colour
7. Q. Find the prob of drawing a queen, a king and a knave (jack) in
that order from a pack of cards in 3 consecutive draws, the
cards drawn not being replaced.
Ans:- A queen can be drawn from a pack of cards in 4C1 = 4 different ways.
So, prob of drawing a queen P(A) = 4c1/52c1 = 4/52 =1/13
After drawing a queen and without being replaced, the total number of cards
reduced to 51. Then prob of drawing king will be P(B) = 4c1/51c1 = 4/51.
Similarly, the prob of drawing a knave P(C) = 4/50.
We want the probability of drawing a queen, king or a knave, without
replacement. i.e. P(ABC). Since the draws are independent, the prob of their
joint occurrence is the same as the product of their individual probabilities.
Therefore, the required probability is
P(ABC) = P(A)xP(B)xP(C) = (4/52)x(4/51)x(4/50) =(4x4x4)/(52x51x50) = 0.00048
8. QUESTIONS
• Q1. Out of the numbers 1 to 120, one number is selected at random. What is the
prob that it is divisible by 8 or 10?
• Ans. Here The Sample space S ={1,2,3,…, 119,120}. Then n(S) =120
Let A denotes the event that the selected number is divisible by 8. That is the
selected number is a multiple of 8.
Then A = {8,16,24,32,40,48,56,64,72,80,88,96,104,112,120} and so n(A) = 15
Similarly, B denotes the event that the number selected is divisible (or a multiple)
of 10. Then B = {10,20,30,40,50,60,70,80,90,100,110,120} and so n(B) = 12
We want the prob. that the card drawn is divisible by 8 or 10.
That is, we require P(AUB)
9. Contd….
P(AUB) = P(A) + P(B) – P(AB), where AB is the event that the elements which are
common to both A and B.
Here AB = {40,80,120} and n(AB) = 3
P(A) = n(A)/n(S), P(B) = n(B)/n(S) and P(AB) = n(AB)/n(S)
Therefore, P(AUB) = 15/120 + 12/120 -3/120
= (15+12-3)/120
= 24/120 = 1/5
i.e., The prob that the selected number is a multiple of 8 or 10 is 1/5
10. More information and results
S = A ONLY + AE + E ONLY + ĀĒ
A ONLY = A – AE = AĒ
E ONLY = E – AE = ĀE
From the figure,
AUE = AĒ + AE + ĀE
Sample space S is the union of
4 mutually exclusive events.
S = ĀĒ + (AĒ + AE + ĀE)
S = ĀĒ + AUE or AUE = S – ĀĒ
And ĀĒ = S - AUE
AĒ A E ĀE
11. Addition theorem for 3 events
A, B, C are 3 events (not necessarily be mutually exclusive) then
P(AUBUC) = P(A) + P(B) + P(C) - P(A∩B) - P(B∩C) - P(A∩C) + P(A∩B∩C) This
can also be written as S
P(AUBUC) = P(A) + P(B) + P(C) - P(AB) - P(BC) - P(AC) + P(ABC)
Note
In addition theorem or union of sets, odd combinations of events
are to be added and even combinations are to be subtracted.
i.e., A,B,C, ABC are to be added and AB,BC,AC are to be
subtracted
A
B
C
12. Qn. Given S = {1,2,3,4,5,6,7,8,9,10}, A = {2,4,6,8,10} B =
{3,5,6,7,8} C = {3,8,6,10}. Establish the result
P(AUBUC) = P(A)+P(B)+P(C)-P(AB)-P(BC)-P(AC)+P(ABC)
ANS: Here, AUBUC ={2,4,6,8,10,3,5,7}
We want to prove if A, B, C are 3 events (not necessarily be mutually exclusive) then
P(AUBUC) = P(A) + P(B) + P(C) - P(AB) - P(BC) - P(AC) + P(ABC)
n(S)=10, n(A)=5, n(B)=4, n(C)=4 & n(AUBUC) = 8
AB={6,8} and n(AB)=2 BC={3,6,8} and n(BC)=3
AC={6,8,10} and n(AC)=3 ABC={6,8} and n(ABC)=2
L.H.S P(AUBUC) = n(AUBUC)/n(S) = 8/10 …….(1)
P(A) = n(A)/n(S) = 5/10. Similarly, P(B)=5/10, P(C)=4/10
P(AB)=2/10, P(BC)=3/10, P(AC)=3/10 and P(ABC)=2/10
R.H.S = P(A) + P(B) + P(C) - P(AB) - P(BC) - P(AC) + P(ABC)
= 5/10 + 4/10 + 4/10 – 3/10 – 3/10 – 3/10 + 2/10
= (5+5+4-2-3-3+2)/10 = 8/10 …………(2)
From (1) & (2), we get LH.S = R.H.S which prove the result.
A
B
C
S
13. EXERCISES
Q1. 17 cards numbered 1,2,3,…,16,17 are placed in an urn and mixed thoroughly. A person
draws a card from it. Find the prob that
a) Its an odd number b) It is divisible by 3 c) A prime
d) It is an even number or divisible by 5 e) Divisible by 2 or 3
Q2. The probabilities of solving a problem by Ramesh and Satheesh are 60% and 72% respectively.
What is the prob that a) the problem is solved b) problem is not solved
Q3. Given that P(A)=0.6, P(E)=0.5, P(AUE)=0.9, then find P(AE), P(ĀĒ), P(Ā), P(AĒ) and P(Ē)
Q4. Prob that a contractor will get a plumbing contract is 2/3. The prob that he will
not get an electric contract is 5/9. If the prob of getting at least one contract is
4/5.What is the prob that he will get both the contracts?
Q5. Prob that Mr. Tom passes a maths exam is 2/3 and that of statistics exam is 4/9. The prob
that he passing at least one subject is 7/9. Find the prob that
i) he will pass both the subjects ii) he fails in statistics
iii) he fails in maths iv) he fails in both subjects
