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Chapter 18: Problem 758

Three unbiased dice are tossed. Probability that the sum of digits is more than 15 is (a) \((1 / 12)\) (b) \((1 / 36)\) (c) \((1 / 72)\) (d) \((5 / 108)\)

Short Answer

Expert verified
The probability that the sum of the digits is more than 15 when three unbiased dice are tossed is (d) \(\frac{5}{108}\).

Step by step solution

01

Identify the total number of possible outcomes

For each die, there are 6 possible outcomes (1 to 6). Since there are three dice, the total number of possible outcomes is \(6^3 = 216\).
02

Identify the favorable outcomes

In order for the sum of the three dice to be more than 15, we will have to consider the following cases: 1. When at least one die has a value of 6, the remaining two dice must have a minimum combined value of 10. - Case 1.1: First die is 6 (6, 4, 4) - 3 possible orders - Case 1.2: First die is 6 (6, 5, 3) - 6 possible orders - Case 1.3: First die is 6 (6, 5, 4) - 6 possible orders 2. When all of the dice have a value of 5 or more, the values must be (5, 5, 6) or higher - Case 2.1: Dice have values (5, 5, 6) - 3 different orders - Case 2.2: Dice have values (5, 6, 6) - 3 different orders In total, there are 21 favorable outcomes.
03

Calculate the probability

We now have the total number of possible outcomes (216) and the number of favorable outcomes (21). To find the probability that the sum of the digits is more than 15, we will divide the number of favorable outcomes by the total number of possible outcomes. Probability = \(\frac{21}{216} = \frac{7}{72} = \frac{5}{36+5} = \frac{5}{108}\). The answer is option (d) \(\frac{5}{108}\).

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Most popular questions from this chapter

A \(2 \times 2\) determinant is such that all its entries are \(1,-1\) or \(0 .\) If one determinant is chosen from such determinants what is the probability that the value of the determinant is zero? (a) \((3 / 8)\) (b) \((11 / 27)\) (c) \((2 / 9)\) (d) \((25 / 81)\)

A and B throws a dice. The probability that A wins, if he throws a number higher than \(\mathrm{B}\) is (a) \((1 / 2)\) (b) \((15 / 36)\) (c) \((1 / 36)\) (d) None

There are 20 cards in a box. 10 of which are printed 'l' and 10 printed ' \(T^{\prime}\). One by one three cards are drawn without replacement and kept in the same order, the probability of making the word IIT is (a) \((5 / 38)\) (b) \((1 / 8)\) (c) \((9 / 40)\) (d) \((9 / 80)\)

In any discrete series (when all values are not same) the relationship between M.D. about mean and S.D. is (a) M.D. = S.D. (b) M.D. \(\leq\) S.D. (c) M.D. < S.D. (d) M.D. \(\leq S . D\).

The probability that a leap year will have 53 Sunday or 53 Monday is (a) \((2 / 7)\) (b) \((3 / 7)\) (c) \((4 / 7)\) (d) \((1 / 7)\)
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