Ques. 1. Two dice are thrown. What is the probability of getting a) the sum of the numbers of both dice are greater than 9? b) Sum is divisible is by 5 c) A doublet d) One is odd
Answers
Answer:
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Step-by-step explanation:
a/c)Scoring a sum greater than 9 or a doublet is =
(4,6),(6,4),(5,5),(5,6),(6,5),(6,6),(1,1),(2,2),(3,3),(4,4)
total ways = 10
The probability of scoring a sum greater than 9 or a doublet is =10/36=5/18
d)To get an odd number, you need to get an even number on one die, and an odd number on the other.
If you roll the first die, and it’s even, you have a 50% chance of an odd number on the second.
If you roll the first die and it’s odd, you have a 50% chance of an even number on the second.
It doesn’t matter what the first die rolls, you have a 50% chance of rolling the correct kind of number (even or odd) on the second die.
b)Here are all the possible outcomes when a pair of dice is thrown :
(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 ,total number of outcomes = 36
Now , number of favourable outcomes (sum divisible by 5) = (1,4)(2,3)(3,2)(4,6)(5,5)(6,4)
= 6
Probability of getting a sum divisible by 5 = Number of favourable outcomes/ total number of outcomes
= 6/36
= 1/6
c)Doublet is the sum of no.'s on both the dice greater than 9 which has been answered in a).
d)To get an odd number, you need to get an even number on one die, and an odd number on the other.
If you roll the first die, and it’s even, you have a 50% chance of an odd number on the second.
If you roll the first die and it’s odd, you have a 50% chance of an even number on the second.
It doesn’t matter what the first die rolls, you have a 50% chance of rolling the correct kind of number (even or odd) on the second die.
Answer:
a)1/4 (or 25%)
b)7/12 (or 58.3%)
c)1/2 (or 50%)
d)1/4 (or 25%)
Step-by-step explanation:
a) We know that a dice has six faces and the numbers on each face are 1,2,3,4,5,6.
The possible combinations for the sum of the numbers to be greater than 9:
4+6,5+6,5+5,6+6.
So, there are 4 possible combinations.
In order for that, the first dice thrown must be 4,5 or 6. The probability of that is 3 out of 6 numbers, which is:
3/6=1/2 (or 50%)
The second dice requires more logical thinking.
If the first dice thrown is 4, the second dice must be 6. If the first dice thrown is 5, the second dice can be 6 or 5. If the first dice thrown is 6, the second dice can be 6,5 or 4.
So, there are 1+2+3=6 possibilities in each scenario. The probability of that is 6 out of 12 numbers (because you thrown a second time), which is:
6/12=1/2 (or 50%)
The probability of getting the sum of the numbers of both dice are greater than 9:
1/2*1/2
=1/4 (or 25%)
b) The multiples of 5 up to 15:
5,10,15
Possible combinations for the sum of the numbers to be divisible by 5:
1+4,2+3,6+4,5+5
So there are 4 possible combinations.
In order for that, the first dice thrown can be any number. The probability of that is 6 out of 6 numbers, which is:
6/6=1 (or 100%)
Similarly, the second dice requires more thinking.
If the first dice thrown is 1, the second dice must be 4. If the first dice thrown is 2, the second dice must be 3. If the first dice thrown is 3, the second dice must be 2. If the first dice thrown is 4, the second dice can be 1 or 6. If the first dice thrown is 5, the second dice must be 5. If the first dice thrown is 6, the second dice must be 4.
So, there are 1+1+1+2+1+1=7 possibilities in each scenario. The probability of that is 7 out of 12 numbers, which is:
7/12 (or 58.3%)
The probability of getting the sum of the numbers of both dice is divisible by 5:
1*7/12
=7/12 (or 58.3%)
c) Definition of doublets: A situation when two dices are thrown at once and both of them ends up getting same digits
Possible combinations of doublets:
1 and 1, 2 and 2, 3 and 3, 4 and 4, 5 and 5, 6 and 6
So, there are 6 possible combinations.
In order for that, the first dice thrown can be any number. The probability of that is 6 out of 6 numbers, which is:
6/6=1 (or 100%)
If the first dice thrown is 1, the second dice must be 1. If the first dice thrown is 2, the second dice must be 2. If the first dice thrown is 3, the second dice must be 3. The same situation applies to 4, 5 and 6.
So, there are 1+1+1+1+1+1=6 possibilities in each scenario. The probability of that is 6 out of 12 numbers, which is:
6/12=1/2 (or 50%)
The probability of getting a doublet is:
1*1/2=1/2(or 50%).
Actually, this is relatively easier.
d) Possible combinations for one of them to be odd:
1 and 1,2,3,4,5,6, 3 and 1,2,3,4,5,6, 5 and 1,2,3,4,5,6
Or you can say: 1,2,3,4,5,6 and 1,3,5
In order for that, the first dice can be any number. The probability of that is 1 (or 100%).
The second dice can only be 1,3 or 5 (because this is fixed).
The probability of that is 3 out of 12 numbers, which is:
3/12=1/4 (or 25%)
Thank you for asking such a meaningful question. I apologize if my explanation is too long. I just wanted to maximize the details.