Question

if two dice are thrown what is the probability that the sum is neither 7 nor 11 ​

Answer 1

Answer:

Let us do it stepwise.

Since two dice are thrown, the total number of outcomes in sample space =6×6=36

Event A:- First let us find the probability that the sum is 7 :— (1,6);(2,5);(3,4);(4,3);(5,2);(6,1)=6/36

Event B:-Now, let us find the probability that the sum is 11 :— (5,6);(6,5)=2/36

Since the events getting sum 7 and getting sum 11 are independent of each other and mutually exclusive. Hence,

P(AnB)=P(A)P(B)=0....eq(1)

Hence, the probability that either A or B or both will happen is

P(AUB)=P(A)+P(B)−P(AnB)=P(A)+P(B)

[bcz eq(1)]

So, P(AUB)=6/36+2/36=8/36

Now, as all the required things are with us, let us answer your question…

Now, the probability that the sum is neither 7 nor 11 is

P(AUB)′=1−P(AUB)=1–8/36=28/36=7/9=0.7777

Hope it helps…

Answer 2

Answer:

Number of total outcomes with two dices are n(S)=36

Event of getting a sum of 7

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

n(A) = 6

Event of getting a sum of 11

B = {(5,6),(6,5)}

n(B)=2

(A∩B)==>n(A∩B)=0

A and B are independent (mutually exclusive) events.

P(A∪B)=P(A)+P(B)=636+226=836=29

Probability of getting neither 7 nor 9

P(A U B)= 1-P(AUB) = 1−29=79