No.

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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, ... , e_p} ⇒ 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.