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Nội dung chính Show- When two dice are rolled what is the probability of getting same number on both?Sample QuestionsWhat is the probability of throwing 2 dice simultaneously?How many outcomes are there when 2 dice are thrown simultaneously?When two dice are thrown simultaneously What is the probability of getting two numbers whose product is even?What is the probability of getting two numbers whose product is not even?
Explanation;
Two dice are thrown.
∴ n(S) = 36.
Getting two numbers whose product is even, i.e., one of the two numbers must be even.
Let sự kiện A: Getting even number on first dice.
sự kiện B: Getting even number on second dice.
∴ n(A) = 18, n(B) = 18, n(A ∩ B) = 9
∴ Required probability = P(A ∩ B)
= `("n"("A") + "n"("B") - "n"("A" ∩ "B"))/("n"("S"))`
= `(18 + 18 - 9)/36`
= `27/36`
= `3/4`
Two different dice are thrown the same time. Find the probability of getting :
(i) a doublet
(ii) a sum of 8
Solution:
When two dice are thrown simultaneously, the sample space of the experiment is
(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)
So there are 36 equally likely outcomes.
Possible number of outcomes = 36.
(i)Let E be an sự kiện of getting a doublet.
Favourable outcomes = (1,1), (2,2),(3,3), (4,4), (5,5),(6,6)
Number of favourable outcomes = 6
P(E) = 6/36 = 1/6
Probability of getting a doublet is 1/6 .
(ii)Let E be an sự kiện of getting a sum of 8.
Favourable outcomes = (2,6), (3,5), (4,4), (5,3), (6,2)
Number of favourable outcomes = 5
P(E) = 5/36
Probability of getting a sum of 8 is 5/36.
Answer
Verified
Hint: In this question, we are given that two dice are thrown simultaneously and we have to find various probabilities of numbers shown on both dice. For this, we will first make sample space and then use that to find favorable outcomes for finding each probability. Total outcomes will be given as the number of elements in sample space. Probability of any sự kiện is given as $textProbability=dfractextNumber of favorable outcomestextTotal outcomes$.
Complete step by
step answer:
Here, we are given that two dice are thrown simultaneously. As we know, a dice has 6 possibilities, therefore for two dice, the number of possibilities will be $6times 6=36$. Let us draw sample space for the given sự kiện.
(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)
Hence, the total number of outcomes are 36.
With the help of this sample space we will find required elements for every part.
(i) Here we have to find the probability of getting the sum as 8. Therefore, let us analyze the sample space and count the numbers whose sum is 8. As we can see, following are required cases:
(2,6), (3,5), (4,4), (5,3), (6,2)
Hence, the number of favorable outcomes is 5. So,
$textProbability=dfrac536$.
Hence, the probability of getting the sum as 8 is $dfrac536$.
(ii) Let us analyze the sample space and count numbers whose one number is multiple of 2 i.e. 2, 4, 6 and another number is multiple of 3 i.e. 3, 6. As we can see, following are required cases:
(2,3), (2,6), (4,3), (4,6), (6,3), (6,6), (3,2), (3,4), (3,6), (6,2), (6,4).
Hence, the number of favorable outcomes is 11. So,
$textProbability=dfrac1136$.
Hence, probability of getting a multiple of 2 on one dice and multiple of 3 on other dice is $dfrac1136$.
(iii) Now, let us count numbers whose sum is least 10, therefore, we have to count numbers whose sum is 10, 11 or 12. As we can see, following are the required cases:
(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)
Hence, the number of favorable outcomes are 6. So,
$textProbability=dfrac636=dfrac16$.
Hence, the probability of getting a sum least 10 is $dfrac16$.
Note: Students should carefully count all the possibilities while calculating probability. In (ii) part, students should note that multiple of 2 or 3 can be on any of the two dices. For example, (2,3) and (3,2) both are favorable cases. In (iii) part, students should note that sum should be least 10, so, they have to consider sum as 10 or higher. Try to avoid mistakes while making sample space.
Probability is a measure of the possibility of how likely an sự kiện will occur. It is a value between 0 and 1 which shows us how favorable is the occurrence of a condition. If the probability of an sự kiện is nearer to 0, let’s say 0.2 or 0.13 then the possibility of its occurrence is less. Whereas if the probability of an sự kiện is nearer to 1, lets say 0.92 or 0.88 then it is much favourable to occur.
Probability of an sự kiện
The probability of an sự kiện can be defined as a number of favorable outcomes upon the total number of outcomes.
P(A) = Number of favorable outcomes / Total number of outcomes
Some terms related to probability
- Experiment: An experiment is any action or set of action performed to determine the probability of an sự kiện. The result of action performed is random or uncertain. e.g. Tossing a coin, rolling dice, etc.Event:
An sự kiện can be defined as certain condition which can happen while performing an experiment. e.g. getting head while tossing a coin, getting even number while rolling dice, etc.Sample Space: It is set of all the possible outcomes which happens after performing an experiment. e.g. Sample Space of tossing a coin = H,T and Sample Space of rolling a dice = 1,2,3,4,5,6, so on.Sample Point: It is a part of sample space which contains one of the
outcomes from Sample Space. e.g. Getting 1 while rolling dice, getting an ace of Spades while drawing a card from pack of cards, etc.Types of Events: There are majorly four kinds of events that are-
- Complimentary events- It is used to find probability of not happening of an sự kiện. It is denoted by ( ‘ ) symbol. If sự kiện is denoted by A, then complimentary of sự kiện is A’. e.g. probability of not getting 2 while rolling a dice. It can be calculated
by subtracting normal probability from 1 i.e. P(A’) = 1 – P(A)Impossible sự kiện- Impossible sự kiện is a type of sự kiện which can never happen. The probability of Impossible sự kiện is 0. e.g. getting a number 8 while rolling a dice.Certain sự kiện- Certain sự kiện is a type of sự kiện which always happen. The probability of a certain sự kiện is 1. e.g. getting a head or a tail after tossing a coin.Equally Likely
events- Events whose probability of occurrence are equal i.e. they are equally likely to happen. The value of probability of such events are same. e.g. getting a head and getting a tail both have 50% probability.
When two dice are rolled what is the probability of getting same number on both?
Since, the number of outcomes while rolling a dice = 6
Number of outcomes while rolling two dice = 62
= 36
The Sample Space for rolling a die is given as,
(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)
Sample points of getting same number on both dice- (1,1) ,(2,2) ,(3,3) ,(4,4) ,(5,5) & (6,6).
Thus, the number of favourable outcomes = 6
Total number of outcomes = 36
P (getting same number on both dice) = 6/36
= 1/6
Hence, the probability of getting same number on both the dice is 1/6.
Sample Questions
Question 1: Find the probability of getting odd number on first dice and even number on other dice when two dice are thrown simultaneously.
Answer:
Total number of outcomes = 36
Sample Space :
(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)
In 9 outcomes we will get odd number on first dice and even number on second dice.
So, required probability is 9/36 = 1/4
Question 2: If two dice are thrown together then find the probability of getting 1 or 2 on either of the dice.
Answer:
Total number of outcomes = 36
Sample Space :
(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)
From above sample space it is clear that there are total 20 possibilities in which 1 or 2 appears on either of the dice.
So, required possibility = 20/36 = 5/9
Question 3: In an sự kiện 2 dice are thrown simultaneously. Find the probability of getting prime number on first dice.
Answer:
Total number of possibilities = 36
Sample Space :
(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)
Since, 2, 3 and 5 are prime number which appear on first dice in 2nd, 3rd and 5th row of sample space respectively.
So, number favourable sample points = 18
Required probability = 18/36
Question 4: Three coins are tossed together find the probability of getting least one head and one tail.
Answer:
Number of possibilities while tossing a coin = 2
Number of possibilities while tossing 3 coins together = 23
= 8
Sample Space :
(H,H,H) , (H,H,T) , (H,T,H) , (H,T,T) ,
(T,H,H) , (T,H,T) , (T,T,H) , (T,T,T)
6 sample points are having head and tail both.
P(E) = 6/8
= 3/4
Question 5: Find the probability of getting least two tails when a coin is tossed three times.
Answer:
Total number of outcomes = 8
Sample Space :
(H,H,H) , (H,H,T) , (H,T,H) , (H,T,T) ,
(T,H,H) , (T,H,T) , (T,T,H) , (T,T,T)
Number of favourable outcomes = 4
Probability = 4/8
= 1/2
What is the probability of throwing 2 dice simultaneously?
If getting the same number on both dice is taken as sự kiện, Then for the 1st sự kiện, ⇒ Number of outcomes are 6. ∴ The probability of getting the same number of both the dice is 1/6.How many outcomes are there when 2 dice are thrown simultaneously?
Note that there are 36 possibilities for (a,b). This total number of possibilities can be obtained from the multiplication principle: there are 6 possibilities for a, and for each outcome for a, there are 6 possibilities for b. So, the total number of joint outcomes (a,b) is 6 times 6 which is 36.When two dice are thrown simultaneously What is the probability of getting two numbers whose product is even?
So, the probability of getting two numbers whose product is even =2736=34.What is the probability of getting two numbers whose product is not even?
∴ The probability of getting two numbers whose product is odd is 1/4. Tải thêm tài liệu liên quan đến nội dung bài viết Two dice are thrown simultaneously what is the probability of getting two numbers