Concept explainers
(a)
To find:
(a)
Answer to Problem 102E
The probability is 0.1667.
Explanation of Solution
Given:
Number of dices rolled in a toss = 2
Number of sides in a die = 6
Formula used:
The probability can be calculated with the formula mentioned below
The rule of complements states
Calculation:
The sample space on rolling a pair of dice is
1 | 2 | 3 | 4 | 5 | 6 | |
1 | 1,1, | 1,2 | 1,3 | 1,4 | 1,5 | 1,6 |
2 | 2,1 | 2,2 | 2,3 | 2,4 | 2,5 | 2,6 |
3 | 3,1 | 3,2 | 3,3 | 3,4 | 3,5 | 3,6 |
4 | 4,1 | 4,2 | 4,3 | 4,4 | 4,5 | 4,6 |
5 | 5,1 | 5,2 | 5,3 | 5,4 | 5,5 | 5,6 |
6 | 6,1 | 6,2 | 6,3 | 6,4 | 6,5 | 6,6 |
Total number of outcomes
The outcomes with doubles
So, the probability of rolling doubles is
(b)
To find: Probability of rolling doubles on the second toss but not on the first.
(b)
Answer to Problem 102E
The probability is
Explanation of Solution
Calculation:
From the sub part a,
Probability of rolling doubles
Using complements rule as mentioned above, the probability of not rolling dice is
Therefore, the required probability
(c)
To find: The probability of first double occurring in the third toss.
(c)
Answer to Problem 102E
The required probability is
Explanation of Solution
Calculation:
From the above sub parts
So, the probability of a double occurring on the third toss but not on the first and second will be
Thus, the probability is 0.1157.
(d)
To find: The probability of occurring doubles on the kth toss.
(d)
Answer to Problem 102E
Explanation of Solution
Let X be the number of tosses for doubles to occur
Probability of doubles occurring in the fourth toss
Probability of doubles occurring in the fifth toss
It can be seen that each time the result is being multiplied by
This follows the general rule
Chapter 5 Solutions
The Practice of Statistics for AP - 4th Edition
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