# 10th Class Mathematics Probability

Probability

Category : 10th Class

Probability

Probability

1. Probability: Probability is the quantitative measure of the degree of certainty of the occurrence of an event.

1. The probability of a sure event is 1

1. The probability of an impossible event is 0.

1. The probability of an event E is a number P (e) such that $0\le P(E)\,\,\,\le \,\,1$

1. A random experiment: An operation which produces some well-defined outcome such that all possible outcomes are known but the exact outcome is unpredictable is called a random experiment.

1. Event: The collection of all or some of the possible outcomes of a random experiment is called an event.

1. Probability of occurrence of an event: The probability of occurrence of a event E is given by:

$P(E)\,=\,\,\frac{Number\,of\,outcomes\,favourable\,to\,event\,\,E}{Total\,number\,of\,possible\,outcomes}$

1. Elementary Event: An event having only one outcome is called an elementary event.

The sum of the probabilities of all the elementary events of an experiment is 1

1. For any event E, $\mathbf{P}\left( \mathbf{E} \right)\mathbf{+}\,\,\overline{\mathbf{E}}\,\,\mathbf{= 1}$, where stands for ‘not E’. E and $\overline{E}$ are called complementary events.

1. Probability of an event cannot be negative and lies between 0 and 1.

1. In a pack of 52 cards we have:

 (i)  4 suits - spades, hearts, clubs and diamonds having 13 cards each. (ii) Each suit has one ace, one king, one queen, one jack and 9 other cards from 2 to 10. (iii) King, queen and jack are called face cards or picture cards. (iv) Hearts and diamonds are red coloured cards while spades and clubs are black coloured cards.

Snap Test

1. 1000 tickets of a lottery were sold and there are 5 prizes on these tickets. If Mahir has purchased one lottery ticket, what is the probability of winning a prize?

(a) $\frac{2}{200}$       (b) $\frac{1}{200}$

(c) $\frac{1}{100}$       (d) $\frac{2}{100}$

(e) None of these

Ans.     (b)

Explanation: Total number of all possible outcomes = number of tickets = 1000

Number of favourable outcomes for winning a prize = number of tickets bearing prizes = 5

$\therefore$    Probability of winning a prize =$\frac{5}{1000}=\frac{1}{200}$

1. A card is drawn from an ordinary pack and a gambler bets that it is a spade or an ace. What are the odds against his winning this bet?

(a) $\frac{4}{9}$           (b) $\frac{9}{5}$

(c) $\frac{9}{4}$           (d) $\frac{9}{2}$

(e) None of these

Ans.     (c)

Explanation: Total number of all possible outcomes = Total number of cards in the pack = 52.

Let E be the event of the gambler winning the bet i.e. getting a spade or an ace.

There are 13 cards of spades and 3 more aces.

$\therefore$ Number of outcomes favourable to the event $E\text{ }=\text{ }13\text{ }+\text{ }3\text{ }=\text{ }16$ and the number of outcomes not favourable to the event $E\text{ }=\text{ }52\text{ }-\text{ }16\text{ }=\text{ }32.$

So, the odds against the gambler winning the bet =$\frac{36}{16}=\frac{9}{4}$.

1. One card is drawn from a well-shuffled deck of 52 cards. Calculate the probability that the card drawn is not an ace.

(a) $\frac{4}{13}$         (b) $\frac{1}{13}$

(c) $\frac{3}{26}$         (d) $\frac{12}{13}$

(e) None of these

Ans.     (d)

Explanation: P (card drawn is not an ace)

= 1 - P (card drawn is an ace)

$=\,\,\,1-\frac{4}{52}\,\,=\,\,\frac{48}{52}\,\,=\,\,\frac{12}{13}.$

1. In a simultaneous throw of two coins, the probability of getting at least one head is:

(a) $\frac{3}{4}$                     (b) $\frac{1}{3}$

(c) $\frac{2}{3}$                       (d) $\frac{1}{2}$

(e) None of these

Ans.     (a)

Explanation: All possible outcomes are HH, HT, TH asd TT.

Let E be the event of getting at least one head.

The outcomes favourable to the event E are HH, HT and TH.

Number of favourable outcomes = 3.

$\therefore$    P (getting at least one head) $=\text{ }P\left( E \right)\text{ }=~\,\,\frac{3}{4}$

1. What is the probability of getting a sum 9 from two throws of a dice?

(a) $\frac{1}{6}$           (b) $\frac{1}{8}$

(c) $\frac{1}{9}$           (d) $\frac{1}{12}$

(e) None of these

Ans.     (c)

Explanation: Total number of outcomes when a dice is thrown twice = 36.

$\therefore$      Total number of all possible outcomes = 36.

Let E be the event of getting a sum 9.

The outcomes favourable to the event E are $\left( 3,\text{ }6 \right),\text{ }\left( 4,\text{ }5 \right),\text{ }\left( 5,\text{ }4 \right)\text{ }and\text{ }\left( 6,\text{ }3 \right)$.

Number of favourable outcomes = 4.

$\therefore$     $P\left( getting\text{ }a\text{ }sum\text{ }9 \right)\text{ }=\text{ }P\text{ }\left( E \right)\,\,=\,\,\,\frac{4}{36}\,\,\,\,=\,\,\,\,\frac{1}{9}$ .

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