Two unbiased dice are thrown, then the probability that the sum is 8 or greater appears on the first dice is ____.

Question

Infinity Learn · Accepted Answer

It is given in the question that two unbiased dice are thrown. So, the total possible outcomes are

6^{2} = 36

.
Let’s calculate all the possible total outcomes for getting accurate results without any assumption.
{(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)}
Let us consider that P (A) is the probability of getting a sum of both dice faces 8 or greater than 8. From all the total possible outcomes mentioned about this is all the possible outcomes for A:
{(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) }
P (A) =

\frac{15}{36}

=

\frac{5}{12}

Let us consider that P (B) is the probability of getting 4 on the first dice.
From all the total possible outcomes mentioned about this is all the possible outcomes for B :
{(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)}
P (B) =

\frac{6}{36} = \frac{1}{6}

Let us consider P(A∩B) is the probability of getting 4 on the first die and the sum greater than or equal to 8.
Sample space (A∩B)={(4,4),(4,5),(4,6)}
P (A∩B)=

\frac{3}{36} = \frac{1}{12}

The probability that sum of the numbers is greater than or equal to 8 given that 4 was thrown on first dice is:

P (\frac{A}{B}) = \frac{P (A \cap B)}{P (B)}

.
Substituting the values,

P (\frac{A}{B}) = \frac{(\frac{1}{12})}{(\frac{1}{6})}

.
⇒

\frac{1}{2}

Two unbiased dice are thrown, then the probability that the sum is 8 or greater appears on the first dice is ____.

Detailed Solution

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