Modern Definition of Probability : Set Theoretic Approach : Problems Solutions

No. Problems & Solutions [Click Solution link to bring up the solutions page]
01.
If S denotes the sample space, then P(S) is

Solution Hide/Show
[1]

P(S) denotes the probability of occurrance of the sample space.
⇒ Probability of occurrence of at least one of the sample points

This is a certainty. Thus P(S) = 1.

Explanation

Consider the sample space consisting of "p" elementary events

Where "S" represents the sample space i.e. the set of all elementary events of the experiment.

S = {e1, e2, e3, ... , ep} ⇒ n(S) = p

Probability of occurrence of the sample space ⇒ P(S) =
n(S)
n(S)
= 1

02.
If sample space contains 10 equally likely simple events, then the probability of each simple event is

Solution Hide/Show
[1/10]

If there are 'n' elementary events (sample points) in an experiment all of which are equally likely and mutually exclusive, the probability of occurrence of each elementary event is equal to 1/n.

Since there are 10 elementary events (sample points) in the experiment, the probabilty of occurrence of each elementary (sample point) is 1/10.

Explanation

Consider the sample space consisting of the 10 elementary events

Where "S" represents the sample space i.e. the set of all elementary events of the experiment.

S = {e1, e2, e3, ... , e10} ⇒ n(S) = 10

Let E1, E2, E3, ... , E10 represent the events each with an elementary event as its sample points.

Thus, E1 = {e1} ⇒ n(E1) = 1
Probability of occurrence of each elementary event ⇒ P(E1) =
n(E1)
n(S)
=
1
10

Similarly, n(E2) = 1, n(E3) = 1, ..., n(E10) = 1 and P(E2) = P(E3) = ... = P(E10) =
1
10

03.
The probability of an event containing 5 sample points in a sample space of 15 sample points is.

Solution Hide/Show
[1/3]

If there are 'n' elementary events (sample points) in an experiment all of which are equally likely and mutually exclusive, the probability of occurrence of each elementary event is equal to 1/n.

Since there are 15 elementary events (sample points) in the experiment, the probabilty of occurrence of each elementary (sample point) is 1/15.

Let "A" be the event consisting of the 5 sample points.
The probabilty of occurrence of the event = Sum of the probabilities of each of the sample points
P(A) =
1
15
+
1
15
+
1
15
+
1
15
+
1
15
=
1 + 1 + 1 + 1 + 1
15
=
5
15
=
1
3

Explanation

Consider the sample space consisting of the 15 elementary events

Where "S" represents the sample space i.e. the set of all elementary events of the experiment.

S = {e1, e2, e3, ... , e15} ⇒ n(S) = 15

Let E1, E2, E3, ... , E15 represent the events each with an elementary event as its sample points.

Thus, E1 = {e1} ⇒ n(E1) = 1
Probability of occurrence of each elementary event ⇒ P(E1) =
n(E1)
n(S)
=
1
10

Similarly, n(E2) = 1, n(E3) = 1, ..., n(E15) = 1 and P(E2) = P(E3) = ... = P(E15) =
1
15

Let "A" be the event consisting of the 5 sample points.
The probabilty of occurrence of the event = Sum of the probabilities of events with each of the sample points
P(A) = P(E1) + P(E4) + P(E12) + P(E5) + P(E9)
=
1
15
+
1
15
+
1
15
+
1
15
+
1
15
=
1 + 1 + 1 + 1 + 1
15
=
5
15
=
1
3

04.
S is a sample space. S = { x n/1 < x ≤ 100} and E = {x / (x+1) (x-1) s}. then P(E) =

Solution Hide/Show
[1/11]

"S", the sample space is defined as {x n/1 < x ≤ 100}

⇒ S = {x n/1 < x ≤ 100}

⇒ S = {2, 3, 4, ... , 100} ⇒ n(S) = 99

Event "E", is defined as {x / (x+1) (x-1) s}

⇒ E = {x / (x+1) (x-1) s}

⇒ E = {2, 3, ... 10} ⇒ n(E) = 9

The probabilty of occurrence of the event "E"
⇒ P(E) =
n(E)
n(S)
=
9
99
=
1
11

05.
If p(A) = 0, then A = Φ

Solution Hide/Show
[TRUE]

Where "A" represents an event of the experiment whose sample space is "S".
⇒ P(A) =
n(A)
n(S)

Given P(A) = 0
Therefore,
n(A)
n(S)
= 0

⇒ n(A) = 0

Therefore, A = φ

06.
If s is a sample space containing 8 elements then the number of probability functions of s with co – domain { 0, 1} is

Solution Hide/Show
[]

The event

07.
Describe explicity the sample spaces for each of the following experiments:
(i) The tossing of four coins.
(ii) The throwing of three dice.
(iii) The tossing of ten coins with the aim of observing the numbers of tails coming up.
(iv) Two cards are selected from a standard deck of cards.
(v) Four successive draws (a) with replacement and (b) with out replacement, from bag containing fifty colored balls out of which ten are white, twenty blue and twenty red.
(vi) A survey of families with two children is conducted and the sex of the children (the older child first) is recorded.
(vii) A survey of families with three children is made and the sex of the children (in order of age, oldest child first) are recorded.
(viii) Three distinguishable objects are distributed in three numbered cells.
(ix) A poker hand (five cards) is dealt from an ordinary deck of cards.
(x) Selecting r screws from the lot produced by a machine, a screw can be defective or non-defective.

Solution Hide/Show
[]

The event

07.
Describe sample space appropriate in each of the following cases:-
(i) n- Tosses of a coin with head or tails as out come in each toss.
(ii) Successive tosses of a coin until a head turns up.
(iii) A survey of families with two children’s is conducted and the sea of the children (the older child first) is recorded.
(iv) Two successive draws (a) with replacement (b) with out replacement from a bag containing 4 colored toys ut of which one is white, one black and 2 red toys.

Solution Hide/Show
[Head, Tail]

The event

08.
Suppose an experiment has n out comes, A1, A2 ---------- An and that it is repeated r times. Let x1, x2 -------- xn record the number of occurrence of A1, A2 ---------- An. describe the sample space. Show that the number of sample points is.

Solution Hide/Show
[Head, Tail]

The event

09.
If A is an event of a sample space S, then the odds in favour of A are

The probability of sample space is

Solution Hide/Show
[Head, Tail]

The event

10.
Let a sample space be S = [ a1, a2, a3]. Which of the following defines probability space on S?

i) P(a1) = 1/4, P(a2) = 1/3, P(a3) =1/3
ii) P(a1) = 2/3, P(a2) = - 1/3, P(a3) = 2/3
iii) P(a1) = 0, P(a2) = 1/3, P(a3) = 2/3.

Solution Hide/Show
[Head, Tail]

The event

11.
A manufacturer buys parts from four different vendor’s numbered 1, 2, 3 and 4. referring to orders placed on two successive days (1,4) denotes the events that on the first day, the order was given to vendor 1 and on the second day it was given to vendor 4. letting A represent the event that vendor 1 gets at least one of these two orders, B the event that the same vendor gets at least one of these two orders, B the event that the same vendor gets both orders and C the event that vendors 1 and 3 do not get either order.

List the element of a) entire sample space (b) A (c) B (d) C (e) A (f) A (g) B ∪ C (h) A ∩ B
(i) A ∩ C (j) A ∪ B and (k) A − B .

Solution Hide/Show
[Head, Tail]

The event

12.
Let s be the samplw space and let A and B be the events. Find an expression for the events.

i) A or B occurs
ii) Weather A nor B occurs
iii) Only A (but not B) occurs.

Solution Hide/Show
[Head, Tail]

The event

13.
Let S = { 1, 1/2, ... ... ... (1/2)n } be a classical event space and A, B be events given by A = {1, 1/2} B = {(1/2)K / k is an even positive integer}. Find p(AB)

Solution Hide/Show
[Head, Tail]

The event

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No. Problems for Practice
01. A coin is tossed three times in succession, the number of sample points in the sample space is?
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