Science > Mathematics > Statistics and Probability > Probability > Problems Based on Numbered Cards
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 cards.
Drawing a Single Numbered Card / Ticket:
Example – 01:
Tickets numbered from 1 to 20 are mixed up together and then a ticket is drawn at random what is the probability of getting a ticket bearing
Solution:
The sample space is S = {1, 2, 3, …….., 20}.
One ticket is drawn at random.
Hence n(S) = 20C1 = 20
a) an even number
Let A be the event of getting ticket bearing an even number
Favourable points are 2, 4, 6, 8, 10, 12,14, 16, 18, 20
∴ n(A) = 10C1 = 10
By the definition P(A) = n(A)/n(S) = 10/20 = 1/2
Therefore the probability of getting ticket bearing even number is 1/2
b) an odd number
Let B be the event of getting ticket bearing an odd number
Favourable points are , 3, 5, 7, 9, 11, 13, 15, 17, 19
∴ n(B) = 10C1 = 10
By the definition P(B) = n(B)/n(S) = 10/20 = 1/2
Therefore the probability of getting ticket bearing odd number is 1/2
c) a perfect square
Let C be the event of getting ticket bearing a perfect square
Favourable points are 1, 4, 9,16
∴ n(C) = 4C1 = 4
By the definition P(C) = n(C)/n(S) = 4/20 = 1/5
Therefore the probability of getting ticket bearing a perfect square is 1/5
d) multiple of four (or divisible by four)
Let D be the event of getting ticket bearing a number multiple of 4
Favourable points are 4, 8, 12, 16, 20
∴ n(D) = 5C1 = 5
By the definition P(D) = n(D)/n(S) = 5/20 = 1/4
Therefore the probability of getting ticket bearing a number multiple of 4 is 1/4
e) multiple of three (or divisible by three)
Let E be the event of getting ticket bearing a number multiple of 3
Favourable points are 3, 6, 8, 12, 15, 18
∴ n(E) = 6C1 = 6
By the definition P(E) = n(E)/n(S) = 6/20 = 3/10
Therefore the probability of getting ticket bearing a number multiple of 3 is 3/10
f) a number greater than 4
Let F be the event of getting ticket bearing a number greater than 4
Favourable points are 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20
∴ n(F) = 17C1 = 17
By the definition P(F) = n(F)/n(S) = 17/20
Therefore the probability of getting ticket bearing a number greater than 4 is 17/20
g) a number less than 11
Let G be the event of getting ticket bearing a number less than 11
Favourable points are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
∴ n(G) = 10C1 = 10
By the definition P(G) = n(G)/n(S) = 10/20 = 1/2
Therefore the probability of getting ticket bearing a number less than 11 is 1/2
h) perfect square or less than 5
Let H be the event of getting ticket bearing a number a perfect square or less than 5
Favourable points are 1, 4, 9, 16, 2, 3
∴ n(H) = 6C1 = 6
By the definition P(H) = n(H)/n(S) = 6/20 = 3/10
Therefore the probability of getting ticket bearing a perfect square or less than 5 is 3/10
i) a prime number
Let J be the event of getting ticket bearing a number a prime number
Favourable points are 2, 3, 5, 7, 11, 13, 17, 19
∴ n(J) = 8C1 = 8
By the definition P(J) = n(J)/n(S) = 8/20 = 2/5
Therefore the probability of getting ticket bearing a prime number is 2/5
j) a prime number or a perfect square
Let K be the event of getting ticket bearing a number a prime number or a perfect square
Favourable points are 2, 3, 5, 7, 11, 13, 17, 19, 1, 4, 9, 16
∴ n(K) = 12C1 = 12
By the definition P(K) = n(K)/n(S) = 12/20 = 3/5
Therefore the probability of getting ticket bearing a prime number or perfect square is 3/5
k) an even number or a perfect square
Let L be the event of getting ticket bearing an even number or a perfect square
Favourable points are 2, 4, 6, 8, 10, 12,14, 16, 18, 20, 1, 9
∴ n(L) = 12C1 = 12
By the definition P(L) = n(L)/n(S) = 12/20 = 3/5
Therefore the probability of getting ticket bearing a prime number or perfect square is 3/5
l) an even number or a number divisible by 5
Let M be the event of getting ticket bearing an even number or a number divisible by 5
Favourable points are 2, 4, 6, 8, 10, 12,14, 16, 18, 20, 5, 15
∴ n(M) = 12C1 = 12
By the definition P(M) = n(M)/n(S) = 12/20 = 3/5
Therefore the probability of getting ticket bearing a prime number or perfect square is 3/5
m) a perfect square or a number multiple of 3
Let N be the event of getting ticket bearing a perfect square or a number multiple of 3
Favourable points are 1, 4, 9, 16, 3, 6, 12,15, 18
∴ n(N) = 9C1 = 9
By the definition P(N) = n(N)/n(S) = 9/20
Therefore the probability of getting ticket bearing a perfect square or a number multiple of 3 is 9/20
n) a number greater than 9 and an even number.
Let Q be the event of getting ticket bearing a number greater than 9 and an even number
Favourable points are 10, 12, 14, 16, 18, 20
∴ n(Q) = 6C1 = 6
By the definition P(Q) = n(Q)/n(S) = 6/20 = 3/10
Therefore the probability of getting ticket bearing a number greater than 9 and an even number is 3/10
o) an even number and multiple of 3
Let R be the event of getting ticket bearing an even number and multiple of 3
Favourable points are 6, 12, 18
∴ n(R) = 3C1 = 3
By the definition P(R) = n(R)/n(S) = 12/20 = 3/5
Therefore the probability of getting ticket bearing a prime number or perfect square is 3/5
In the next article, we shall study some basic problems of probability based on drawing two or more cards from the collection of numbered cards.