Probability

**Category : **10th Class

**Probability**

**Probability **

** **

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

**The probability of a sure event is 1**

** **

**The probability of an impossible event is 0.**

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

**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.

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

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

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

**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

**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.**

**Probability of an event cannot****be negative and lies between 0 and 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**

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

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

** **

**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)

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

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

** **

**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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