Two perfect dice are thrown. What is the probability that the sum of the numbers on the faces is <strong>NEITHER</strong> 9 <strong>NOR</strong> 10?

Explanation

The total number of outcomes is 36. The outcomes for a sum of 9 are (3,6), (4,5), (5,4), (6,3), which is 4 cases. The outcomes for a sum of 10 are (4,6), (5,5), (6,4), which is 3 cases. The probability of getting 9 or 10 is \frac{4+3}{36} = \frac{7}{36}. Therefore, the probability of getting neither is 1 - \frac{7}{36} = \frac{29}{36}.

Related questions on Statistics & Probability

• Let x be the mean of squares of first n natural numbers and y be the square of mean of first n natural numbers. If $\frac{x}{y}=\fra...
• What is the probability of getting a composite number in the list of natural numbers from 1 to 50?
• Two numbers x and y are chosen at random from a set of first 10 natural numbers. What is the probability that (x+y) is divisible by 4?
• A number x is chosen at random from first n natural numbers. What is the probability that the number chosen satisfies $x+\frac{1}{x} \gt...
• Three fair dice are tossed once. What is the probability that they show different numbers that are in AP?