If two dice are tossed simultaneously, what is the probability of the sum of the numbers on the dice being (a) an even number (c) a two digit number (b) equal to 10
Answers
Answer:
a) 1,3,5 ,7,11,13,17,19,23,27,29,31,33,even number
total number 36
probability is
13/36
b) equal to 10
4+6,6+4,5+5,
total 3
sum of all probability× 36
probability. 3/36
c) 2 digit number. 27
total probability. 36
total probability. 27/36
Answer:
Step-by-step explanation:
a) If two dice are tossed simultaneously ,then total number of outcomes = 6² = 36.
Required probability = 1 - ( the probability of the getting the sum of the numbers on the two dice is an even number ).
Favourable cases to get the sum an even number = ( 2,2),(2,4), (2,6), (4,6), (6,6).
Probability of the getting the sum of the numbers on the two dice is an even number = 5/6.
∴ The required probability = 1 - 5/6 = 6-5/6= 1/6.
b) If two dice are tossed simultaneously, then total number of outcomes = 6² = 36.
Required probability = 1 - ( the probability of the getting the sum of the numbers on the two dice is equal to 10) .
Favourable cases to get the sum equal to 10 = ( 4,6), (5,5), (6,4)..
Probability of the getting the sum of the numbers on the two dice is an equal to 10= 3/6 = 1/2.
∴ The required probability = 1 - 1/2 = 2-1/2 = 1/2.
c) If two dice are tossed simultaneously, then total number of outcomes = 6² = 36.
Required probability = 1 - ( the probability of the getting the sum of the numbers on the two dice is a two digit number ).
Favourable cases to get the sum a two digit number = ( 4,6), (5,5), ( 5,6), (6,4), ( 6,5), (6,6).
Probability of the getting the sum of the numbers on the two dice is a two digit number = 6/6 = 1.
∴ The required probability = 1 - 1 = 0.
P (E) = 0 ,so,E is an impossible event.