Sometimes we say, “probably it may rain” or “Probably he may get more than 90% in the examination" etc. These are elements of certainty. Means we are not certain about some things. In mathematics these comes under Probability.
The theory of probability is widely used in the area of natural as well as social science.
Probability as a Measure of Uncertainty
Suppose we through a die which is a well balanced cube with its six faces marked numbers from 1 to 6, one number of one face, we see the number which come up on its uppermost face. A die can fall with any of its face upper most.
The number on each of the face is equally libely and possible outcome. There are six equally likely ealy outcomes: 1, 2, 3, 4, 5 or 6 in a single throw of a die. The chance of any number ‘say 3’ to come up is 1 out of 6. That is probability of 3 coming up is 1/6
i.e. p(3) = 1/6
Similarly in tossing a coin, we may get either head (H) or tail (T) up and P(H) = ½
Hence probability of an event E is
There are only six possible outcome in a single throw of a die. If we want to find probability of 7 or 8 to come up, then in that case number of possible or favorable outcome is O (zero), hence P(7) = 0/6 = 0
i.e. probability of an impossible event is zero.
If we consider to find the probability of number less than 7, then all six cases are favorable and hence P(number less than 7) = 6/6 = 1
i.e. probability of sure event is 1
Now, P (3) = 1/6, then probability of numbers of other than 3 must be 5/6
Example 1. An unbiased dice is tossed.
- Write the sample space of the experiment
- Find the probability of getting a number greater than 4.
- Find the probability of getting a prime number.
Solution:-
- Sample space = {1, 2, 3, 4, 5, 6}
n(s) = 6
- E = event of getting a number greater than 4
= {5, 6}
n (E) = 2
P (> 4) = Probability of a number greater than 4
= n(E)/n(S) = 2/6 = 1/3
- E = Event of getting a prime number
= {2, 3, 5}
n (E) = 3
P(Prime number) = Probability of a prime number
Example 2. A bag contains 5 red balls, 8 White balls, 4 green balls and 7 black balls. A ball is drawn at random from the bag. Fine the probability that it is. (i) Black (ii) not green.
Solution:- Red balls = 5
White balls = 8
Green balls = 4
Black balls = 7
Total balls = 24
(i) P (Black balls) = 7/24
(ii) P (not a green ball) = 1- P (green ball)
= 1 - 4/24 = 20/24
= 5/6
Example 3. A card is drawn at random from a pack of 52 playing cards. Find the probability that the card drawn is neither an ace nor a king.
Solution:- P (neither an ace nor a king)
= 1 – p (either an ace or a king)
= 1 – 8/52 {no. of ace = 4}
{no. of king = 4}
Total = 8
= (52 – 8)/52
= 44/52
= 11/13
Example 4. Out of 400 bulbs in a box, 15 bulbs a defective. One ball is taken out at random from the box. Find the probability that the drawn bulb is not defective.
Solution:- Total number of bulbs = 400
Total number of defective bulb = 15
Total number of non-defective bulbs = 400-15
= 385
P (not defective bulb) = 385/400
= 77/80
Example 5. Find the probability of getting 53 Fridays in a leap year.
Solution:- No. of days in a leap year = 366
366 days = 52 weeks and 2 days.
A leap year must has 52 Fridays
The remaining two days can be
(a) Sunday an Monday
(b) Monday and Tuesday
(c) Tuesday and Wednesday
(d) Wednesday and Thursday
(e) Thursday and Friday
(f) Friday and Saturday
(g) Saturday and Sunday
Out of 7 case, 2 cases have Friday
P (53 Friday) = 2/7
Example 6. Three unbiased coins are tossed simultaneously. What is the probability of getting exactly two heads?
Solution: - When three coins are tossed simultaneously, the sample space is
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}.
n(S) = 8
E = Set of cases favorable to the event
= {HHT, HTH, THH}
n(E) = 3
P (exactly two heads) =
Example 7. A dice is thrown twice. Find the probability of getting (a) doublets (b) prime number on each die.
Solutions: - Sample space =
S = { (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) }
n (S) = 36
(i) E = Events getting doublet
= {(1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6)}
n (E) = 6
P(doublet) =
(ii) E = Events getting prime number on each die.
= {(2, 2), (2, 3), (2, 5), (3, 2), (3, 3), (3, 5), (5, 2), (5, 3), (5, 5)}
n(E) = 9
P (getting prime number on each die)
= n(E)/n(S) = 9/36 = 1/4
Exercise - 31
- (1) A dice is thrown once. Find the probability of getting
(a) A number greater than 3
(b) A number less than 5 - A bag contain 5 black, 7 red and 3 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is
(a) Red
(b) Black or white
(c) Not black - A bag contains 4 red 5 black and 6 white balls. A ball is drawn from the bag a random. Find the probability that the ball drawn is
(a) White
(b) Red
(c) Not black
(d) Red or white - A card is drawn at random from a pack of 52 playing cards. Find the probability that the card drawn is neither a queen nor a jack.
- Tickets numbered from 1 to 20 are mixed up together and then a ticket is drawn at random. What is the probability that the ticket has a number which is a multiple of 3 or 7?
- In a single throw of dice, what is the probability of
(a) An odd number on one dice and 6 on the other
(b) A number greater than 4 on each dice
(c) A total of 11
(d) Getting same number on either dice. - A die is thrown twice. Find the probability of getting
(a) doublets
(b) number greater than 5 on one dice. - Three coins are tossed simultaneously. Find the probability of getting
(a) Exactly 2 heads
(b) No heads - In a simultaneous toss of four coins, What is the probability of getting:
(a) Less than 2 heads?
(b) Exactly 3 head
(c) More than 2 heads? - Three coins are tossed once. Find the probability of:
(a) 3 heads
(b) exactly 2 heads
(c) at least two heads
Answers
1. 1/2, 2/3, | 2. 7/15, 8/15, 2/3 | 3. 2/5, 4/15, 2/3, 2/3 | 4. 11/13, |
5. 2/5 | 6. 1/6, 1/9, 1/18, 1/6 | 7. 1/6, 11/36 | 8. 3/8, 1/8 |
9. 5/6, 3/8, 5/16 | 10. 1/8, 3/8, 1/2 |
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We are already familiar with plotting a point on a plane graph paper. For this we take two perpendicular lines XoX’ and YoY’ intersecting at O. XOX’ is called x-axis or abscissa and YoY’ is called y-axis or ordinate.
Point in a plane
Let us take a point P in a plane. Let XOX’ and YOY’ be pendicualr to each other at O. are drawn. If OM = x and ON = y then x-coordinate of P is x and y-coordinate of P is y. Here we write x-coordinate first. Hence (x, y) and (y, x) are different point whenever
.
The two lines XOX’ and YOY’ divides the plane into four parts called quadrants. XOY, YOX’, X’OY’ and Y’OX are respectively the first second, third and and fourth quadrants. We take the direction from O to X and O to Y as positive and the direction from O to X’ and O to Y’ as negative.
Distance between two points
Let P (x1, y1) and Q (x2, y2) be the two points. We have to find PQ.
OM = x1, PM = y1 = RN
ON = x2, QN = y2
PR = MN = ON – OM
= x2 – x1
QR = QN – RN = y2 – y1
By Pythagoras theorem
PQ2 = PR2 + QR2
= (x2 - x1)2 + (y2 – y1)2
If x1 = 0, y1 = 0, x2 = x and y2 = y
Then
Section Formula
Let P (x, y) divided a line AB such that AP : PB = m1 : m2.
Let coordinates of A are (x1, y1) and B are (x2, y2).
It is obvious that
Taking,
Similarity
Note
(i) if P is mid point of AB, then AP : PB = 1 : 1
(ii) If m1 : m2 = k, then coordinates of P are
Example 1. Show that the points A (5, 6), B (1, 5), C (2, 1) and D (6, 2) are the vertices of a square.
Solution: - Using Distance Formula
Diagonal AC = diagonal BD
A, B, C and D are vertices of a square
Example 2. Determine the ratio in which the point P (m, 6) divides the join of A(-4, 3) and B(2, 8). Also find the value of m.
Solution:-
Let AP : PB = K : 1
Co-ordinates of P are
Burt co-ordinates of P are given (m, 6)
Example 3. The coordinates of the mid-point of the line joining the points (2p + 2, 3) and (4, 2q + 1) are (2p, 2q). Find the value of p and q.
Solution:-
Let the given points be A (2p + 2, 3), B(4, 2q + 1) and C (2p, 2a)
Co-ordinates of C are
Example 4. Find the ratio in which the line-segment joining the points (6, 4) and (1, -7) is divided by x-axis.
Solution:-
Let the two points A (6, 4) and B (1, -7), be divided by x-axis at C (x, 0)
Let AC : CB = K : 1
Co-ordinates of C are
As C lies on x-axis
Example 5. Prove that coordinates of the centroid of a triangle ABC, with vertices (x1 y1), (x2, y2) and (x3, y3) are given by
Solution:-
Let the vertices of triangles be A(x1, y1), B(x2, y2) and C(x3, y3)
Let D be the mid-point of BC.
Co-ordinates of D are
Let centroid of triangle ABC be G.
Co-ordinates of G are
Example 6. Find the value of m for which the points with co-ordinates (3, 5), (m, 6) and (1/2, 15/2) are collinear.
Solution :- Let the three points be A (3, 5) B (m, 6) and (1/2, 15,2)
As points A, B and care collinear
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