Two dice are thrown. Find the conditional probability of getting 2 atleast once given that a sum of 7 is obtained. Arrange it in the form of a chart
Answers
Answer:
The conditional probability = 1/3.
Step-by-step explanation:
Given that two dice are thrown, simultaneously or one after another.
The combinations of the digits on the dice that result in the sum of 7 are:
1 + 6, 2 + 5, 3 + 4, 4 + 3, 5 + 2 , and 6 + 1.
Total number of combinations = 6.
Number of combinations that contain the digit 2 = 2.
If digit 2 appears twice, then the sum is not equal to 7.
So the conditional probability, given that the sum is 7 , of getting the digit 2 at least once = 2/6 = 1/3.
<===== answer.
Probability that the sum of two digits on two dice = 7 is
= 6/36 = 1/6
as there are totally 36 outcomes if two dice are thrown and there are 6 outcomes with the sum of 7.
So overall probability that the digit 2 occurs and the sum of 7 is obtained is = 1/6 * 1/3 = 1/18.
Answer:
1/3
Step-by-step explanation:
Conditional probability of event A occurring given event B occurs is given by
P(A|B)=P(A&B)/P(B)
Where P(A&B) is the probability that both events occur SIMULTANEOUSLY
Let A : event when 2 occurs on a die
Let B : event when sum is 7
P(A&B)=2/36
where favourable cases are (2,5) and (5,2)
P(B)=6/36
Where favourable cases are (1,6) (2,5) (3,4 ) (4,3) (5,2) and (6,1)
So required probability=P(A|B)=(2/36)÷(6/36)= 2/6= 1/3
Hope this answer helped you