Math, asked by ReenaGonsalves, 1 month ago

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 ​

Answers

Answered by leehun
3

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.

Answered by tennetiraj86
4

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:-

  • Probability of an event P (E) = Number of favourable outcomes/ Total number of possible outcomes.

  • A die thrown "n" times or "n" dice are thrown simultaneously then the total number of possible outcomes are 6^n.
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