Problems Based on Numbered Cards 02

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In the last article, we have studied problems on the throwing of dice. In this article, we shall study to solve problems to find probability involving numbered tickets.

Drawing Two or More Numbered tickets:

Example – 01:

Tickets numbered from 1 to 50 are mixed up together and then two tickets are drawn at random what is the probability that

Solution:

The sample space is S = {1, 2, 3, …….., 50}.

Two tickets are drawn at random.

Hence n(S) = ⁵⁰C₂ = 1225

a) both the tickets bear an even number

Let A be the event that both the tickets bear an even number

Favourable points are 2, 4, 6, 8, 10, … , 50

There are 25 favourable points

∴ n(A) = ²⁵C₂ =  300

By the definition P(A) = n(A)/n(S) = 300/1225 = 12/49

Therefore the probability that both the tickets bear an even number is 12/49

b) both the tickets bear an odd number

Let B be the event that both the tickets bear an odd number

Favourable points are 21, 3, 5, 7, ….., 49

There are 25 favourable points

∴ n(B) = ²⁵C₂ =  300

By the definition P(B) = n(B)/n(S) = 300/1225 = 12/49

Therefore the probability that both the tickets bear an odd number is 12/49

c) both the tickets bear a perfect square

Let C be the event that both the tickets bear a perfect square

Favourable points are 1, 4, 9, 16, 25, 36, 49

There are 7 favourable points

∴ n(C) = ⁷C₂ =  21

By the definition P(C) = n(C)/n(S) = 21/1225 = 3/175

Therefore the probability that both the tickets bear a perfect square is 3/175

d) Both the tickets bear a number multiple of four (or divisible by four)

Let D be the event that both the tickets bear a number multiple of four

Favourable points are 4, 8, 12, 16, …., 48

There are 12 favourable points

∴ n(D) = ¹²C₂ =  66

By the definition P(D) = n(D)/n(S) = 66/1225

Therefore the probability that both the tickets bear a number multiple of four is 66/1225

e) both the tickets bear a number multiple of three (or divisible by three)

Let E be the event that both the tickets bear a number multiple of three

Favourable points are 3, 6, 9, …., 48

There are 16 favourable points

∴ n(E) = ¹⁶C₂ =  120

By the definition P(E) = n(E)/n(S) = 120/1225 = 24/245

Therefore the probability that both the tickets bear a number multiple of three is 24/245

f) both the tickets bear a number greater than 44

Let F be the event that both the tickets bear a number greater than 44

Favourable points are 45, 46, 47, 48, 49, 50

There are 6 favourable points

∴ n(F) = ⁶C₂ =  15

By the definition P(F) = n(F)/n(S) = 15/1225 = 3/245

Therefore the probability that both the tickets bear a number greater than 44 is 3/245

g) both the tickets bear a number less than 11

Let G be the event that both the tickets bear a number less than 11

Favourable points are 1, 2, 3, ……, 10

There are 10 favourable points

∴ n(G) = ¹⁰C₂ =  45

By the definition P(G) = n(G)/n(S) = 45/1225 = 9/245

Therefore the probability that both the tickets bear a number less than 11 is 9/245

h) both the tickets bear perfect square or number less than 10

Let H be the event that both the tickets bear perfect square or number less than 10

Favourable points are 1, 4, 9, 16, 25, 36, 49, 2, 3, 5, 6, 7, 8

There are 13 favourable points

∴ n(H) = ¹³C₂ =  78

By the definition P(H) = n(H)/n(S) = 45/1225 = 9/245

Therefore the probability that both the tickets bear perfect square or a number less than 5 is 78/1225

i) both the tickets bear a prime number

Let J be the event that both the tickets bear a prime number

Favourable points are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47

There are 15 favourable points

∴ n(J) = ¹⁵C₂ =  105

By the definition P(J) = n(J)/n(S) = 105/1225 = 3/35

Therefore the probability that both the tickets bear a prime number is 3/35

j) both the tickets bear a prime number or a perfect square

Let K be the event that both the tickets bear a prime number or a perfect square

Favourable points are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37,

41, 43, 47, 1, 4, 9, 16, 25, 36, 49

There are 22 favourable points

∴ n(K) = ²²C₂ =  231

By the definition P(K) = n(K)/n(S) = 231/1225 = 33/175

Therefore the probability that both the tickets bear a prime number or a perfect square is 33/175

both the tickets bear a number greater than 35 and an even number.

Let L be the event that both the tickets bear a number greater than 35 and an even number

Favourable points are 36, 38, 40, 42, 44, 46, 48, 50

There are 8 favourable points

∴ n(L) = ⁸C₂ =  28

By the definition P(L) = n(L)/n(S) = 28/1225 = 4/175

Therefore the probability that both the tickets bear a number greater than 35 and an even number is 4/175

k) both the tickets bear an even number and multiple of 5

Let M be the event that both the tickets bear an even number and multiple of 5

Favourable points are 10, 20, 30, 40, 50

There are 5 favourable points

∴ n(L) = ⁵C₂ =  10

By the definition P(M) = n(M)/n(S) = 10/1225 = 2/245

Therefore the probability that both the tickets bear an even number and multiple of 5 is 2/245

In the next article, we shall study some basic problems of probability based on the drawing of a single card from a pack of playing cards.

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