Some Special Sample Spaces
Probability of Class 10
A die is thrown once |
S = {1,2,3,4,5,6}; n(S) = 6 |
A coin is tossed once |
S{H,T}; n(S) = 2 |
A coin is tossed twice or Two coins are tossed simultaneously |
S{HH,HT,TH,TT}; n(S) = 4 = 2² |
A coin is tossed three times or Three coins are tossed simultaneously |
S{HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}; n(S) = 8 = 2³ |
Two dice are thrown together or A die is thrown twice |
S{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6), (2,1),(2,2),(2,3),(2,4),(2,5),(2,6), (3,1),(3,2),(3,3)(3,4),(3,5),(3,6) (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)} n(S) = 6²
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question 1. A die is thrown once. What is the probability of getting a prime number?
Solution: In a single throw of a die, all possible outcomes are 1, 2, 3, 4, 5, 6.
Total number of possible outcomes = 6.
Let E be the event of getting a prime number.
Then, the favourable outcomes are 2, 3, 5.
Number of favourable outcomes = 3.
∴ P(getting a prime number) = P(E) = 3/6 = 1/2
question 2. A coin is tossed once. What is the probability of getting a head?
Solution: When a coin is tossed once, all possible outcomes are H and T.
Total number of possible outcomes = 2.
The favourable outcome is H.
Number of favourable outcome = 1
∴ P(getting a head)
= 1/2
question 3. A bag contains a red ball, a blue ball and a yellow ball, all the balls being of the same size. Kritika takes out a ball from the bag without looking into it. What is the probability that she takes out the
(i) Yellow ball?
(ii) Red ball?
(iii) Blue ball?
Solution: Kritika takes out a ball from the bag without looking into it. So, it is equally likely that she takes out any one of them.
Let Y be the event ‘the ball taken out is yellow’, B be the event ‘the ball taken out is blue’, and R be the event ‘the ball taken out is red’.
Now, the number of possible outcomes = 3.
(i) The number of outcomes favourable to the event Y = 1.
So, P(Y) = 1/3
Similarly, (ii) P(R) = 1/3 and (iii) P(B) = 1/3
question 4. All the three face cards of spades are removed from a well-shuffled pack of 52 cards. A card is then drawn at random from the remaining pack. Find the probability of getting.
(i) Black face card (ii) a queen (iii) a black card.
Solution: After removing three face cards of spades (king, queen, and jack) from a deck of 52 playing cards, there are 49 cards left in the pack. Out of these 49 cards one card can be chosen in 49 ways.
∴ Total number of elementary events = 49
(i) There are 6 black face cards out of which 3 face cards of spades are already removed. So, out of remaining 3 black face cards one black face card can be chosen in 3 ways.
∴ Favourable number of elementary events = 3
Hence, P (getting a black face card) = 3/49
(ii) There are 3 queens in the remaining 49 cards. So, out of these three queens, one queen can be chosen in 3 ways
∴ Favourable number of elementary events = 3
Hence, P (Getting a queen) = 3/49
(iii) There are 23 black cards in the remaining 49 cards. So, out of these 23 black cards, one black card can be chosen in 23 ways
∴ Favourable number of elementary events = 23
Hence, P (Getting a black card) = 23/49
question 5. A die is thrown, Find the probability of
(i) prime number (ii) multiple of 2 or 3 (iii) a number greater than 3
Solution: In a single throw of die any one of six numbers 1,2,3,4,5,6 can be obtained. Therefore, the tome number of elementary events associated with the random experiment of throwing a die is 6.
(i) Let A denote the event “Getting a prime no”. Clearly, event A occurs if any one of 2,3,5 comes as outcome.
∴ Favorable number of elementary events = 3
Hence, P (Getting a prime no.) = 3/6 = 1/2
(ii) An multiple of 2 or 3 is obtained if we obtain one of the numbers 2,3,4,6 as out comes
∴ Favorable number of elementary events = 4
Hence, P (Getting multiple of 2 or 3) = 4/6 = 2/3
(iii) The event “Getting a number greater than 3” will occur, if we obtain one of number 4,5,6 as an outcome.
∴ Favorable number of outcomes = 3
Hence, required probability = 3/6 = 1/2
question 6. Two unbiased coins are tossed simultaneously. Find the probability of getting
(i) two heads (ii) at least one head (iii) at most one head
Solution: If two unbiased coins are tossed simultaneously, we obtain any one of the following as an outcome:
HH, HT, TH, TT
∴ Total number of elementary events = 4
(i) Two heads are obtained if elementary event HH occurs.
∴ Favorable number of events = 1
Hence, P (Two heads) = 1/4
(ii) At least one head is obtained if any one of the following elementary events happen :
HH, HT, TH
∴ favorable number of events = 3
Hence P (At least one head) = 3/4
(iii) If one of the elementary events HT, TH, TT occurs, than at most one head is obtained
∴ favorable number of events = 3
Hence, P (At most one head) = 3/4
question 7. Two coins are tossed simultaneously. What is the probability of getting at least one head?
Solution: When two coins are tossed simultaneously, all possible outcomes are HH, HT, TH, TT.
Total number of possible outcomes = 4.
Let E be the event of getting at least one head.
Then, the favourable outcomes are HT, TH, HH.
Number of favourable outcomes = 3.
∴ P(getting at least one head) = P(E) = 3/4.
question 8. One card is drawn at random from a well-shuffled deck of 52 cards. What is the probability that the card drawn is a face card?
Solution: Total number of all possible outcomes = 52.
We know that kings, queens and jacks are the face cards.
Number of face cards = 12.
Let E be the event of getting a face card.
Then, the number of favourable outcomes = 12.
∴ P(getting a face card) = 12/52 = 3/13
question 9. Cards each marked with one of the numbers, 4, 5, 6, …., 20 are placed in a box and mixed thoroughly. One card is drawn at random from the box. What is the probability of getting an even prime number?
Solution: Out of 17 (= 20 − 3) cards, one card can be drawn in 17 ways.
∴ Total number of possible outcomes = 17
There are 6 prime numbers, namely, 5, 7, 11, 13, 17, 19
Here, there is no even prime number.
∴ Favourable number of outcomes out of 17 = 0
Hence, P(getting an even prime number) = 0/17 = 0