Question

Two dice are rolled, write the sample space 'S' and number of sample points N(s). Also write events and number of samples points in the event according to given conditions:
1)sum of the digits on upper face is prime numbers.
2)sum of the digits on on upper face is multiple of 5.
3)sum of the digits on upper face is 25.
4) digits on the upper face of the first die is less then the digits on the second die.

Answer 1

Answer:

THIS QUESTION TAKES MORE TIME TO WRITE ON KEYPAD THEREFORE I HAD WRITING IN MY NOTES.

Answer 2

Step-by-step explanation:

Given:Two dice are rolled.

To find: Write the sample space 'S' and number of sample points N(s). Also write events and number of samples points in the event according to given conditions:

1) Sum of the digits on upper face is prime numbers.

2) Sum of the digits on on upper face is multiple of 5.

3) Sum of the digits on upper face is 25.

4) Digits on the upper face of the first die is less then the digits on the second die.

Solution:

Sample space for rolling two dice simultaneously.

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)}

Number of sample points N(S)=36

1)Sum of the digits on upper face is prime numbers.

Solution: We know that prime numbers are 2,3,5,7,11 upto counting of 12,as maximum sum on 2 dice is 12.

Number of sample points N(S)=36

Sample space for sum of the digits on upper face is prime numbers

C= {(1,1),(1,2),(2,1),(2,3),(3,2),(4,1),(1,4), (1,6),(6,1),(2,5),(5,2),(3,4),(4,3),(5,6),(6,5)}

Number of sample points for sum of the digits on upper face is prime numbers N(C)=15.

Let the Probability of the event is P(S).

$P(S) = \frac{N(C)}{N(S)}$

$P(S) = \frac{15}{36}$

or

$\bf P(S) = \frac{5}{12}$

2) Sum of the digits on upper face is multiple of 5.

Solution: Number of sample points N(s)=36

Sample space for sum of the digits on on upper face is multiple of 5.

A:{(4,1),(1,4),(2,3),(3,2) (4,6),(6,4),(5,5)}

Number of sample points N(A)=7.

Let the Probability of the event is P(A)

$\bf P(A) = \frac{7}{36}$

3) Sum of the digits on upper face is 25.

Solution: Sum of digits on upper face of two dice is 12 maximum.

So, sum of the digits on upper face is 25 can not be possible for two dice.

Thus,

Sample space for this event is B ={ }

Sample points N(B)=0

Let the Probability of sum of the digits on upper face is 25 is P(25),

So,

$P(25) = \frac{0}{36}$

$\bf P(25) = 0$

4) Digits on the upper face of the first die is less then the digits on the second die.

Solution:

Sample space for digits on the upper face of the first die is less then the digits on the second die.

D= {(1,2),(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6),(4,5),(4,6),(5,6)}

Number of sample points N(D)=15

Let the Probability of this event is P(L)

$P(L) = \frac{15}{36}$

$\bf P(L) = \frac{5}{12}$

Final answer:

1) Sum of the digits on upper face is prime numbers=5/12

2) Sum of the digits on on upper face is multiple of 5=7/36

3) Sum of the digits on upper face is 25=0

4) Digits on the upper face of the first die is less then the digits on the second die=5/12

Hope it will help you.

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