If two dice are rolled together, what is the probability that the sum of the dice is less than 7?

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Concept Used:

P(A) = Favorable outcomes/Total number of outcomes

Calculations:

Total number of possible outcomes = 6 × 6 = 36

Favorable outcomes = (1, 6) (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)

⇒ P(A) = 6/36 = 1/6

∴ The probability of getting sum as 7 when two dice are thrown is 1/6.

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Answer

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Hint: Find out total number of outcomes when two dice are thrown simultaneously. Then find the total number of events in which the sum of the numbers on two dice is ‘more than 7’ and ‘less than 7’. Divide the number of desirable outcomes with the total number of outcomes to find the probability of both the cases. Complete step-by-step answer:Since, a cubical dice has six faces and each face has a number that range from 1 to 6. So, the total number of outcomes when 1 dice is thrown is 6. Therefore, when two dice are thrown, the total number of outcomes will be the total combinations of the 6 numbers of one die with the 6 numbers of the other die. Now, let us come to the question. Total number of outcomes or total number of sample space $=n(S)=6\times 6=36$

Combinations in which sum is more than 7 are: (2, 6); (3, 5); (3, 6); (4, 4); (4, 5); (4, 6); (5, 3); (5, 4); (5, 5); (5, 6); (6, 2); (6, 3); (6, 4); (6, 5); (6, 6).Therefore, the total number of outcomes in which sum is more than 7 $=n({{E}_{1}})$$=15$. Combinations in which sum is less than 7 are: (1, 1); (1, 2); (1, 3); (1, 4); (1, 5); (2, 1); (2, 2); (2, 3); (2, 4); (3, 1); (3, 2); (3, 3); (4, 1); (4, 2); (4, 3).Therefore, the total number of outcomes in which sum is less than 7 $=n({{E}_{2}})$$=15$. We know that, probability of an event $=\dfrac{\text{number of desired outcome}}{\text{total number of outcome}}=\dfrac{n(E)}{n(S)}$. Therefore, probability of getting the sum more than 7 $=\dfrac{n({{E}_{1}})}{n(S)}=\dfrac{15}{36}=\dfrac{5}{12}$. Also, probability of getting the sum less than 7 $=\dfrac{n({{E}_{2}})}{n(S)}=\dfrac{15}{36}=\dfrac{5}{12}$. Note: It would be favourable for us to note all the outcomes in a table otherwise we may get confused in counting the desirable number of events. It is important to note that if ‘n’ number of dice is rolled simultaneously then the total number of outcomes will be ${{6}^{n}}$.

Probability means Possibility. It states how likely an event is about to happen. The probability of an event can exist only between 0 and 1 where 0 indicates that event is not going to happen i.e. Impossibility and 1 indicates that it is going to happen for sure i.e. Certainty.

The higher or lesser the probability of an event, the more likely it is that the event will occur or not respectively.

For example – An unbiased coin is tossed once. So the total number of outcomes can be 2 only i.e. either “heads” or “tails”. The probability of both outcomes is equal i.e. 50% or 1/2.
So, the probability of an event is Favorable outcomes/Total number of outcomes. It is denoted with the parenthesis i.e. P(Event).

P(Event) = N(Favorable Outcomes) / N (Total Outcomes)

Note: If the probability of occurring of an event A is 1/3 then the probability of not occurring of event A is 1-P(A) i.e. 1- (1/3) = 2/3

What is Sample Space?

All the possible outcomes of an event are called Sample spaces.

Examples-

A six faced dice is rolled once. So, total outcomes can be 6 and
Sample space will be [1, 2, 3, 4, 5, 6]
An unbiased coin is tossed, So, total outcomes can be 2 and
Sample space will be [Head, Tail]
If two dice are rolled together then total outcomes will be 36 and
Sample space will be [ (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) ]

Types of Events

Independent Events: If two events (A and B) are independent then their probability will be
P(A and B) = P (A ∩ B) = P(A).P(B) i.e. P(A) * P(B)

Example: If two coins are flipped, then the chance of both being tails is 1/2 * 1/2 = 1/4

Mutually exclusive events:

If event A and event B can’t occur simultaneously, then they are called mutually exclusive events.
If two events are mutually exclusive, then the probability of both occurring is denoted as P (A ∩ B) and
P (A and B) = P (A ∩ B) = 0
If two events are mutually exclusive, then the probability of either occurring is denoted as P (A ∪ B) P (A or B) = P (A ∪ B) = P (A) + P (B) − P (A ∩ B) = P (A) + P (B) − 0
= P (A) + P (B)

Example: The chance of rolling a 2 or 3 on a six-faced die is P (2 or 3) = P (2) + P (3) = 1/6 + 1/6 = 1/3

Not Mutually exclusive events: If the events are not mutually exclusive then

P (A or B) = P (A ∪ B) = P (A) + P (B) − P (A and B)

What is Conditional Probability?

For the probability of some event A, the occurrence of some other event B is given. It is written as P (A ∣ B)

P (A ∣ B) = P (A ∩ B) / P (B)

Example- In a bag of 3 black balls and 2 yellow balls (5 balls in total), the probability of taking a black ball is 3/5, and to take a second ball, the probability of it being either a black ball or a yellow ball depends on the previously taken out ball. Since, if a black ball was taken, then the probability of picking a black ball again would be 1/4, since only 2 black and 2 yellow balls would have been remaining, if a yellow ball was taken previously, the probability of taking a black ball will be 3/4.

Solution:

When two dice are rolled together then total outcomes are 36 and Sample space is [ (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) ]
So, pairs with sum 7 are (1, 6) (2, 5) (3, 4) (4, 3) (5, 2) (6, 1) i.e. total 6 pairs
Total outcomes = 36
Favorable outcomes = 6
Probability of getting the sum of 7 = Favorable outcomes / Total outcomes
= 6 / 36 = 1/6
So, P(sum of 7) = 1/6.

Similar Questions

Question 1: What is the probability of getting 1 on both dice?

Solution:

When two dice are rolled together then total outcomes are 36 and Sample space is [ (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) ]
So, pairs with both 1’s are (1,1) i.e. only 1 pair
Total outcomes = 36
Favorable outcomes = 6

Probability of getting pair of 1 = Favorable outcomes / Total outcomes
= 1 / 36
So, P(1,1) = 1/36.

Question 2: What is the probability of getting the sum of 4?

Solution:

When two dice are rolled together then total outcomes are 36 and Sample space is
[ (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) ]
So, pairs with sum 4 are (1,3) (2,2) (3,1) i.e. total 3 pairs
Total outcomes = 36
Favorable outcomes = 3
Probability of getting the sum of 4 = Favorable outcomes / Total outcomes
= 3/36 = 1/12
So, P(sum of 3) = 1/12.

Question 3: What is the probability of getting the sum of 5?

Solution:

When two dice are rolled together then total outcomes are 36 and Sample space is [ (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) ]
So, pairs with sum 5 are (1,4) (2,3) (3,2) (4,1) i.e. total 4 pairs
Total outcomes = 36
Favorable outcomes = 4
Probability of getting the sum of 5 = Favorable outcomes / Total outcomes
= 4 / 36 = 1/9
So, P(5) = 1/9.