Q.Two dice are thrown together. Write the sample space and find the number of sample points. Hence find the probability of getting a sum of 7 on the two dice.

Three six-faced dice give $6^3=216$ sample points.

When three balanced six-faced dice are thrown together, each die can show any of $6$ outcomes, independently of the others. By the multiplication principle the total number of sample points is

$6\times6\times6=6^3=216.$

Measuring the life of an electric bulb gives an infinite sample space, because the lifetime can be any value in a continuous range — option (c).

In the GSEB Class-12 Statistics Probability chapter, a sample space is the set of all possible outcomes of a random experiment.

• Throwing two dice: $36$ outcomes — finite.

• Selecting two employees from an office: a finite number of combinations — finite.

• Selecting a card from 52 cards: $52$ outcomes — finite.

• Measuring the life of an electric bulb: the lifetime can be any positive real number (e.g. any value in hours), so the outcomes are uncountably many — infinite.

Mutually exclusive events cannot occur at the same time; for them $A \cap B = \varnothing$ and $P(A \cap B) = 0$.

GSEB Class-12 Statistics, Probability:

Two (or more) events defined on the same sample space are said to be mutually exclusive (or disjoint) if the occurrence of one prevents the occurrence of the other in the same trial — they have no common outcome.

$A \cap B = \varnothing \quad \Rightarrow \quad P(A \cap B) = 0$

For example, in a single throw of a die, the events 'getting an even number' and 'getting an odd number' are mutually exclusive.

Three six-faced dice give $6\times6\times6=6^3=216$ sample points. Option (D).

Each die can land in $6$ ways, and the three dice are thrown simultaneously (independently). By the fundamental counting principle, the total number of sample points in the sample space is

$6\times6\times6=6^3=216.$

$A$ and $A'$ are disjoint, so $A \cap A' = \varnothing$ and $P(A \cap A') = 0$ — option (b).

The complement $A'$ consists of exactly those sample points that are not in $A$. Hence $A$ and $A'$ can never occur together:

$A \cap A' = \varnothing$

The probability of the impossible (empty) event is zero:

$P(A \cap A') = P(\varnothing) = 0$

(Consistently, $A \cup A' = S$ and $P(A) + P(A') = 1$.)

$S = \{(1,H),\dots,(6,H),(1,T),\dots,(6,T)\}$, containing $6 \times 2 = 12$ outcomes.

The die shows one of $\{1,2,3,4,5,6\}$ and the coin shows one of $\{H, T\}$. Each combined outcome is an ordered pair (die-result, coin-result). By the counting principle there are $6 \times 2 = 12$ outcomes:

$S = \{(1,H),(2,H),(3,H),(4,H),(5,H),(6,H),\ (1,T),(2,T),(3,T),(4,T),(5,T),(6,T)\}$

$n(S) = 12$