10th Class Mathematics Probability

Probability

Probability

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

2. The probability of a sure event is 1

3. The probability of an impossible event is 0.

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

5. 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.

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

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

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

8. 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

9. 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.

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

11. In a pack of 52 cards we have:

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 =\[\frac{5}{1000}=\frac{1}{200}\]

2. 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 =\[\frac{36}{16}=\frac{9}{4}\].

3. 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)

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

4. 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) \[=\text{ }P\left( E \right)\text{ }=~\,\,\frac{3}{4}\]

5. 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.

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