two dies are thrown once find the probability of obtaining (a) a total of 6 (b) a total of 10 (c) a total of more than 9 (d) the same number on both the dies (doublet) (e) sum even number (f) product multiple of 2
2 thoughts on “two dies are thrown once find the probability of obtaining<br />(a) a total of 6<br />(b) a total of 10 <br />(c) a total of more”
Answer:
hope it’s helpful to you
Step-by-step explanation:
The probability of each of these outcomes is 1/36, so the probability of getting a total of 8 is 5/36. The probability of getting an even number on the first die is 1/2.
Answer:
hope it’s helpful to you
Step-by-step explanation:
The probability of each of these outcomes is 1/36, so the probability of getting a total of 8 is 5/36. The probability of getting an even number on the first die is 1/2.
Step-by-step explanation:
Given:–
Two dies are thrown once.
To find:–
Two dies are thrown once find the probability of obtaining
(a) a total of 6
(b) a total of 10
(c) a total of more than 9
(d) the same number on both the dies (doublet)
(e) sum even number
(f) product multiple of 2
Solution:–
Given that
Two dice are thrown once
We know that
A die thrown “n” times or “n” dice are thrown simultaneously then the total number of possible outcomes are 6^n
Nmber of dice = 2
Total number of possible outcomes = 6^2 = 36
They are:
(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).
Rule:–
Probability of an event P (E) =
Number of favourable outcomes/ Total number of possible outcomes
a) The probability of obtaining a total of 6:
Favourable outcomes = (1,5),(2,4),(3,3),(4,2),(5,1)
Number of favourable outcomes = 5
Total number of possible outcomes = 36
Probability of an event P (E) =Number of favourable outcomes/ Total number of possible outcomes
Probability of obtaining a total of 6 on the top face of the dice = 5/36
b) Probability of obtaining a total
of 10:–
Favourable outcomes = (4,6),(5,5),(6,4)
Number of favourable outcomes = 3
Total number of possible outcomes = 36
Probability of an event P (E) =Number of favourable outcomes/ Total number of possible outcomes
Probability of obtaining a total of 10 on the top face of the dice
= 3/36
=1/12
Probability of obtaining a total of 10 on the top face of the dice = 1/12
c) Probability of getting a total of more than 9:–
Favourable outcomes = (4,6),(5,5,),(5,6),(6,4),(6,5),(6,6)
Number of favourable outcomes = 6
Total number of possible outcomes = 36
Probability of an event P (E) =Number of favourable outcomes/ Total number of possible outcomes
Probability of obtaining a total of more than 9 on the top face of the dice
= 6/36
=1/6
Probability of obtaining a total of more than 9 on the top face of the dice = 1/6
d) Probability of getting the same number on both the dies (doublet) :–
Favourable outcomes = (1,1),(2,2),(3,3),(4,4),(5,5),(6,6)
Number of favourable outcomes = 6
Total number of possible outcomes = 36
Probability of an event P (E) =Number of favourable outcomes/ Total number of possible outcomes
Probability of getting the same number on both the dies (doublet)
= 6/36
=> 1/6
Probability of getting the same number on both the dies (doublet) = 1/6
e) Probability of obtaining sum even number :–
Favourable outcomes = (1,1),(1,3),(1,5),(2,2),(2,4),(2,6),(3,1),(3,3),(3,5),(4,2),(4,4),(4,6),(5,1),(5,3),(5,5),(6,2),(6,4),(6,6)
Number of favourable outcomes = 18
Total number of possible outcomes = 36
Probability of an event P (E) =Number of favourable outcomes/ Total number of possible outcomes
Probability of getting the sum an even number on the top face of the dice
= 18/36
= 1/2
Probability of getting the sum an even number on the top face of the dice =1/2
f) Probability of obtaining the product is a multiple of 2:–
Favourable outcomes = (1,2),(1,4),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,2),(3,4),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,2),(5,4),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)
Number of favourable outcomes = 27
Total number of possible outcomes = 36
Probability of an event P (E) =Number of favourable outcomes/ Total number of possible outcomes
Probability of obtaining the product is a multiple of 2 on the top face of the dice
= 27/36
= 3/4
Probability of obtaining the product is a multiple of 2 on the top face of the dice = 3/4
Answer:–
a)Probability of obtaining a total of 6 on the top face of the dice = 5/36
b)Probability of obtaining a total of 10 on the top face of the dice = 1/12
c)Probability of obtaining a total of more than 9 on the top face of the dice = 1/6
d)Probability of getting the same number on both the dies (doublet) = 1/6
e)Probability of getting the sum an even number on the top face of the dice =1/2
f)Probability of obtaining the product is a multiple of 2 on the top face of the dice = 3/4
Used formula:–