Both the tickets drawn may have prime numbers or both composite numbers

Problem 3

In a lottery of 50 tickets numbered 1 to 50, 2 tickets are drawn simultaneously. Find the probability that both the tickets may have prime numbers or may have composite numbers.

Solution

Total number of tickets

= 50

Number of tickets drawn

= 2

Experiment :

Drawing 2 tickets from the 50 tickets

Total Number of Possible Choices

= Number of ways in which 2 tickets can be drawn from the 50 tickets

⇒ n = 50C2
=
50 × 49
2 × 1
= 25 × 49
= 1,225

Let

A : the event of both the tickets having either a prime number or a composite number on it

For Event A

Prime Numbers :

= 15

{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47}

Numbers which have exactly two distinct factors, 1 and itself.

Composite Numbers :

= 34

{4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 50}

Numbers which have more than two distinct factors.

Others : Neither Composite nor Prime

= 1

{1}

Event A can be accomplished in two alternative ways.

EA1 : Drawing both the tickets with prime numbers on them

EA2 : Drawing both the tickets with composite numbers on them

Prime Composite Others Total
Available 15 34 1 50
To Choose
Alternative 1 2 0 0 2
Choices 15C234C01C050C2
Alternative 2 0 2 0 2
Choices 15C034C21C050C2

Number of favorable choices

= Number of ways in which the two tickets can be drawn having both prime numbers or both composite numbers on them

= Number of ways in which 2 tickets with prime numbers on them can be drawn + Number of ways in which 2 tickets with composite numbers on them can be drawn

= (Number of ways in which 2 tickets with prime numbers on them can be drawn from the available 15) + (Number of ways in which 2 tickets with composite numbers on them can be drawn from the available 34)

⇒ mA = mEA1 + mEA2
= (15C2) + (34C2)
=
15 × 14
2 × 1
+
34 × 33
2 × 1
= (15 × 7) + (17 × 33)
= 105 + 561
= 666

Probability of drawing both the tickets having prime numbers or composite numbers.

⇒ Probability of occurrence of Event A

=
Number of Favorable Choices for the Event
Total Number of Possible Choices for the Experiment
⇒ P(A) =
mA
n
=
666
1,225

Odds

Number of Unfavorable Choices

= Total Number of possible choices − Number of Favorable choices

⇒ mAc = n − mA
= 1,225 − 666
= 559

in favor

Odds in Favor of drawing both the tickets having prime numbers or composite numbers.

⇒ Odds in Favor of Event A

= Number of Favorable Choices : Number of Unfavorable Choices

= mA : mAc

= 666 : 559

against

Odds against drawing both the tickets having prime numbers or composite numbers.

⇒ Odds against Event A

= Number of Unfavorable Choices : Number of Favorable Choices

= mAc : mA

= 559 : 666

Odds (alternative)

Probability of non-occurrence of Event A

⇒ P(Ac) = 1 − P(A)
= 1 −
666
1,225
=
1,225 − 666
1,225
=
559
1,225

in favor

Odds in Favor of the cards drawn belonging to the same suit

⇒ Odds in Favor of Event A

= Probability of occurrence of the event : Probability of non-occurrence of the event
=
559
1,225
:
666
1,225
= 559 : 666

against

Odds against the cards drawn belonging to the same suit

⇒ Odds against Event A

= Probability of non-occurrence of the event : Probability of occurrence of the event
=
559
1,225
:
666
1,225
= 559 : 666