CBSE Class 10 Mathematics Probability Notes Set 05

Basic Terms

A random experiment has outcomes that cannot be predicted with certainty, for example tossing a coin or throwing a die. The set of all possible outcomes is the sample space S, and each element is a sample point. An event is any subset of S.
An impossible event has no outcomes (probability 0) and a sure event is S itself (probability 1). A simple event has exactly one outcome; otherwise it is a compound event. Two events are mutually exclusive if they cannot occur together, and events are equally likely if none is expected in preference to the others.

Probability of an Event

For equally likely outcomes, \( P(E) = \frac{n(E)}{n(S)} \) where n(E) is the number of favourable outcomes and n(S) the total number. Always \( 0 \le P(E) \le 1 \). The complement satisfies \( P(\text{not } E) = 1 - P(E) \), and for exhaustive mutually exclusive events the probabilities add to 1.
Addition rule: \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \); for mutually exclusive events \( P(A\cup B)=P(A)+P(B) \). Odds in favour of E are P(E) : P(not E) and odds against are P(not E) : P(E).

Standard Sample Spaces

One coin: 2 outcomes. Two coins: 4 outcomes (HH, HT, TH, TT). Three coins: 8 outcomes. One die: 6 outcomes with 3 even, 3 odd, 3 prime (2, 3, 5). Two dice: 36 outcomes, with sums 2 to 12 occurring 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1 times; doublets 6; a total of 9 occurs 4 times.
Pack of 52 cards: 4 suits of 13, 26 red and 26 black, 12 face cards (6 red, 6 black), 4 kings, 4 aces. A non-leap year has 52 weeks and 1 extra day, so P(53 Tuesdays) = 1/7; a leap year has 2 extra days, so P(53 Sundays) = 2/7. Choosing two cards: use combinations, for example the number of ways to pick two red cards is 26C2 = 325.

Worked Examples

Example 1: three dice are thrown; total at least 6. The 216 outcomes include only 10 with total 3, 4 or 5, so P = 1 − 10/216 = 103/108.
Example 2: probability that both of two selected children are girls from 2 boys and 3 girls: 3 of the 10 equally likely pairs, so P = 3/10.
Example 3: an event E with P(E) = 0.5 and P(F) = 0.3 and P(E and F) = 0.2 has P(E or F) = 0.5 + 0.3 − 0.2 = 0.6.