Two dice, one blue and one grey, are thrown at the same time. Write down all the possible outcomes. What is the probability that the sum of the two numbers appearing on the top of the dice is (i) 8? (ii) 13? (iii) less than or equal to 12?
Step-by-Step Solution
Key Concept: When two dice are thrown, each die can show any number from 1 to 6 independently. Hence the sample space consists of 6 × 6 = 36 equally likely ordered pairs (blue die result, grey die result). Probability of an event = (number of favourable outcomes) / (total number of outcomes).
1. List of all possible outcomes
The ordered pair \((b,g)\) denotes the number on the blue die \(b\) and the number on the grey die \(g\).
$$\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),\
(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),\
(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),\
(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),\
(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),\
(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}$$
Total outcomes \(N = 36\).
2. Probability that the sum is 8
Find all pairs with \(b+g = 8\):
$$\{(2,6),(3,5),(4,4),(5,3),(6,2)\}$$
Number of favourable outcomes \(F = 5\).
$$P(\text{sum}=8) = \frac{F}{N} = \frac{5}{36}$$
3. Probability that the sum is 13
The maximum possible sum with two dice is \(6+6 = 12\). Hence no outcome gives a sum of 13.
$$F = 0 \quad\Rightarrow\quad P(\text{sum}=13) = \frac{0}{36} = 0$$
4. Probability that the sum is \(\le 12\)
Every possible outcome has a sum between 2 and 12 inclusive. Therefore all 36 outcomes satisfy the condition.
$$F = 36 \quad\Rightarrow\quad P(\text{sum}\le 12) = \frac{36}{36} = 1$$
Correct Answer: All possible outcomes: 36 ordered pairs as listed above.\n(i) $P(\text{sum}=8)=\frac{5}{36}$.\n(ii) $P(\text{sum}=13)=0$.\n(iii) $P(\text{sum}\le 12)=1$.