Notes on Probability | Grade 10 > Compulsory Mathematics > Probability | KULLABS.COM

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The probability of happening of an event A is

$$P(A) = \frac{\text{Number of favourable outcomes of event A}}{\text{Number of possible outcomes}}$$

#### Mathematical Probability

If a trial results in 'n' exhaustive, mutually exclusive and equally likely cases and 'm' of them are favourable to the happening of an event E then the probability of happening of E is given by

$$P(E) = \frac{\text{Favourable number of outcomes (cases)}}{\text{Exhaustive number of cases}} = \frac{m}{n}$$

#### Statistical (Empirical) probability

If a trial is repeated a number of times under essential homogeneous and identical conditions, then the value of the ratio of the number of times of the event happens to the number of all possible outcomes as the number of trials becomes indefinitely large is called probability of happening of the event.

A die is a well-balanced cube with its six faces marked with numbers (dots) from 1 to 6. It has a chance of happening events consisting of the single element is $$\frac{1}{6}$$

An unbiased coin has 50% chance of happening the event 'H' and 'T'. i.e. $$P(H) = \frac{1}{2} = P(T)$$.

A pack of card consists of spade (13), Heart (13), Diamond (13) and club (13). So there is chance of happening spade, heart, diamond and club each equal to $$\frac{13}{52} = \frac{1}{4}$$.

#### Equally likely events

Two events are said to be "Equally likely" if one of them can be expected in preference to other.
Probability of impossible events = 0 and probability of sure event = 1.

#### Mutually exclusive events

Two events A and B are said to be a mutually exclusive event, if $$A \cap B = \phi$$. In other words, such events where the occurrence of one event precludes the occurrence of the other event.

Example:Let's consider a simultaneous toss of two coins, then sample space S = {HH, HT, TH, TT } ∴ $$A \cap B = \phi$$.

#### Additive law for mutually exclusive events

If A and B are two mutually exclusive events, then $$P(A \cup B) = P(A) + P(B).$$ If A, B and C are three mutually exclusive events then occurrence of at least one of these three events is $$P(A \cup B \cup C)= P(A) + P(B) + P(C).$$

Independent event: Two events are said to be independent if the occurrence of one does not depend upon the occurrence of other. E.g. on rolling a dice, tossing a coin occurrence of face '5' and occurrence of the head 'H' are independent events.

#### Multiplication law

If two events A and B are independent and P(A) and P(B) are their respective probability then the probability of their simultaneous occurrence is denoted by $$P(A \cap B)$$ and defined as,
$$P(A\;and\;B) = P(A \cap B) = P(A) \times P(B).$$

Remarks:

• If A, B and C are three independent events then the probability of the simultaneous occurrence is given by $$P(A \cap B \cap C) = P(A) \times P(B) \times P(C)$$

1. If A, B and C are three independent events then the probability of the simultaneous occurrence is given by $$P(A \cap B \cap C) = P(A) \times P(B) \times P(C)$$.
2. The probability that the event A will not happen is 1 - P(B). Therefore, the probability that none of the two events A and B will happen is 1- P(B). Therefore, the probability that none of the two events A and B will happen is [ 1 - P(A)] \times [ 1 - P(B) }.
3. The probability that at least one event will happen = $$1 - [ 1 - P(A) ] \times [ 1 - P(B) ]$$

.

### Very Short Questions

Total balls = 3 red + 4 black + 5 white = 12 balls

Let B be the event getting a black ball.

Then,

P(B) = $$\frac {\text{number of black balls}}{\text{total balls}}$$ = $$\frac 4{12}$$ = $$\frac 34$$

$$\therefore$$ P($$\overline{B}$$) = 1 – P(B) = 1 - $$\frac 34$$ = $$\frac 14$$Ans

Total no. of balls = 5 + 3 + 2 = 10

No. of balls other than blue balls = 5 + 2 = 7

Now,

P(not getting a blue ball) = $$\frac 7{10}$$Ans

Here,

n = Number of possible cases = 100

m = Number of favorable cases = Number of integers from 1 to 100 divisible by 3 = 33

$$\therefore$$ P(multiple of 3) = $$\frac mn$$ = $$\frac {33}{100}$$Ans

The prime numbers from 5 to 20 are 5, 7, 11, 13, 17, 19.

$$\therefore$$ number of cards of prime numbers = 6

Total number of cards numbered from 5 to 20 = 16

Now,

The probability of getting a card of prime number,

P(a prime number) = $$\frac {\text{Number of cards of prime number}}{\text{Total number of cards}}$$ = $$\frac 6{16}$$ = $$\frac 38$$Ans

Here,

Number of red face cards = 6

Total no. of cards = 52

Now,

P(a red face card) = $$\frac {\text{Number of red face cards in a pack}}{\text{Total number of cards in a pack}}$$ = $$\frac 6{52}$$ = $$\frac 3{26}$$Ans

The probability that the face turned may be even number only,

P(even number) = $$\frac {\text{the number of even numbered faces in a dice}}{\text{total faces in the dice}}$$ = $$\frac 36$$ = $$\frac 12$$Ans

Here,

P(getting 6) = $$\frac {\text{number of faces numbered 6}}{\text{total number of faces}}$$ = $$\frac 16$$

$$\therefore$$ P(not getting 6) = 1 – P(getting 6) = 1 - $$\frac 16$$ = $$\frac 56$$Ans

The probability that it will show number less than 4 = P(1) + P(2) + P(3) = $$\frac 16$$ + $$\frac 16$$ + $$\frac 16$$ = $$\frac 36$$ = $$\frac 12$$Ans

Here,

Total number of experiments = sum of the frequencies = 186 + 205 + 211 + 187 + 204 + 207 = 1200

• Number of outcomes which are less than 4 = 186 + 205 + 211 = 602
$$\therefore$$ P(number less than 4) = $$\frac {\text{Total number of outcomes}}{\text{Total number of experiments}}$$ = $$\frac {602}{1200}$$ = $$\frac {301}{600}$$Ans
• Number of outcomes greater than 3 = 187 + 204 + 207 = 598
$$\therefore$$ P(number greater than 3) = $$\frac {\text{number of outcomes}}{\text{total no. of experiments}}$$ = $$\frac {598}{1200}$$ = $$\frac {299}{600}$$Ans

The cards numbered 5, 10, 15, 20, 25, 30 are divisible by 5, and cards numbered 7, 14, 21, and 28 are divisible by 7.

Here,

P(divisible by 5) = $$\frac {number\; of\; cards\; that\; are\; divisible\; by\; 5}{total\; number\; of\; cards}$$ = $$\frac 6{30}$$ = $$\frac 15$$

P(divisible by 7) = $$\frac {number\;of\;cards\;that\;are\;divisible\;by\;7}{total\;number\;of\;cards}$$ = $$\frac 4{30}$$ = $$\frac {2}{15}$$

Now,

P(divisible by 5 or 7) = P(divisible by 5) + P(divisible by 7) = $$\frac 15$$ + $$\frac 2{15}$$ = $$\frac {3 + 2}{15}$$ = $$\frac {5}{15}$$ = $$\frac 13$$Ans

The numbers from 1 to 30 which are exactly divisible by 3 are:

3, 6, 9, 12, 15, 18, 21, 24, 27, 30

The numbers from 1 to 30 which are exactly divisible by 7 are:

7, 14, 21, 28

The numbers exactly divisible by both 3 and 7 is only 21.

Here,

n(S) = 30

n(divisible by 3) = 10

n(divisible by 7) = 4

n(divisible by 3 and 7) = 1

Now,

\begin{align*} P(divisible\;by\;3\;or\;7) &= P(divisible\;by\;3) + P(divisible\;by\;7) - P(divisible\;by\;3\;and\;7)\\ &= \frac{n(divisible\;by\;3)}{n(S)} + \frac{n(divisible\;by\;7)}{n(S)} - \frac{n(divisible\;by\;3\;and\;7)}{n(S)}\\ &= \frac{10}{30} + \frac {4}{30} - \frac {1}{30}\\ &= \frac {13}{30}_{Ans}\\ \end{align*}

The set of number from 1 to 20 which are exactly divisible by 3 is:

A = {3, 6, 9, 12, 15, 18}

The set of number from 1 to 20 which are exactly divisible by 4 is:

B = {4, 8, 12, 16, 20}

From 1 to 20, 12 is only the number exactly divisible by both 3 and 4.

i.e. A∩ B = 12

Now,

n(S) = 20

n(A) = 6

n(B) = 5

n(A∩B) = 1

\begin{align*} \therefore P(A\;or\;B) &= P(A) + P(B) - P(A∩B)\\ &= \frac {n(A)}{n(S)} + \frac {n(B)}{n(S)} - \frac {n(A∩B)}{n(S)}\\ &= \frac 6{20} + \frac 5{20} - \frac 1{20}\\ &= \frac {10}{20}\\ &= \frac 12_{Ans}\\ \end{align*}

Total numbers from 5 to 15 = 11

The prime numbers from 5 to 15 are:

5, 7, 11, 13.

$$\therefore$$ P(getting prime) = $$\frac {prime\;numbers\;from\;5\;to\;15}{numbers\;from\;5\;to\;15}$$ = $$\frac 4{11}$$

Even numbers from 5 to 15 are:

6, 8, 10, 12, 14

$$\therefore$$ P(getting even) = $$\frac {even\; numbers\;between\;5\;to\;15}{numbers\;from\;5\;to\;15}$$ = $$\frac 5{11}$$

Here, the events are mutually execlusive.

Hence,

P(getting prime or even numbers) = P(prime numbers) + P(even numbers) = $$\frac 4{11}$$ + $$\frac 5{11}$$ = $$\frac {9}{11}_{Ans}$$

Here,

n(S) = 52

n(Queen) = 4

n(Black ace) = 2

Hence,

The probability of getting a queen or black ace is:

\begin{align*} P(queen\;or\;black\;ace) &= P(Queen) + P(Black\; ace)\\ &= \frac {n(Queen)}{n(S)} + \frac {n(Black\; ace)}{n(S)} \\ &= \frac 4{52} + \frac 2{52}\\ &= \frac {4 + 2}{52}\\ &= \frac {6}{52}\\ &= \frac {3}{26}_{Ans}\\ \end{align*}

Suppose, B denotes the event getting black cards and F denotes the event getting face cards.

$$\therefore$$ P(B) = $$\frac {number\;of\;black\;cards\;in\;a\;pack}{number\;of\;cards\;in\;a\;pack}$$ = $$\frac {26}{52}$$ = $$\frac 12$$

and P(F) =$$\frac {number\;of\;face\;cards\;in\;a\;pack}{number\;of\;cards\;in\;a\;pack}$$ = $$\frac {12}{52}$$ = $$\frac 3{13}$$

Then,

The probability getting not a faced card:

P($$\overline {F}$$) = 1 - $$\frac 3{13}$$ = $$\frac {10}{13}$$

Again,

The events B and $$\overline {F}$$ are not mutually exclusive events.

So;

P(B∩$$\overline{F}$$) =$$\frac {number\;of\;black\;non\;faced\;cards\;in\;a\;pack}{number\;of\;cards\;in\;a\;pack}$$ = $$\frac {20}{52}$$ = $$\frac 5{13}$$

By the formula,

P(B∪$$\overline{F}$$) = P(B) + P($$\overline {F}$$) -P(B∩$$\overline{F}$$) = $$\frac 12$$ + $$\frac {10}{13}$$ - $$\frac 5{13}$$ = $$\frac {13+20-10}{26}$$ = $$\frac {23}{26}_{Ans}$$

Here,

Total number of balls = 4 black + 3 white = 7 balls

Now,

The probability of getting first black ball;

P1(Black) = $$\frac {number\;of\;black\;balls}{total\;number\;of\;balls}$$ = $$\frac 47$$

If the first ball is drawn and not replaced, the probability of getting second black ball,

P2 (Black) = $$\frac {number\;of\;remaining\;black\;balls}{total\;number\;of\;remaining\;balls}$$ = $$\frac 36$$ = $$\frac 12$$

Therefore,

P(black ball/black ball) = P1 (black ball)× P2 (black ball) = $$\frac 47$$× $$\frac 12$$ = $$\frac 27$$Ans

Total number of balls = 6 + 4 + 5 = 15

Probability of getting first ball is red,

P1 (red) = $$\frac 6{15}$$ = $$\frac 25$$

Probability of getting second ball is white,

P2 (white) = $$\frac 4{15}$$, since first ball is replaced.

Probability of getting third ball is blue,

P3 (blue) = $$\frac 5{15}$$ = $$\frac 13$$, since first and second ball is replaced.

∴ P (red, white, blue) = P1 (red) ×P2 (white) ×P3 (blue) = $$\frac 25$$ ×$$\frac 4{15}$$ ×$$\frac 13$$ = $$\frac 8{255}$$Ans

Here,

Number of kings in a pack of cards = 4

Total no. of cards in the pack = 52

P(a king) = $$\frac {number\;of\;kings\;in\;a\;pack\;of\;cards}{total\;number\;of\;cards\;in\;a\;pack\;of\;cards}$$ = $$\frac 4{12}$$ = $$\frac 13$$

Again,

Number of blue marbles = 3

Total number of marbles = 2 red + 3 blue = 5

P(a blue marble) = $$\frac {number\;of\;blue\;marbles}{total\;number\;of\;marbles}$$ = $$\frac 35$$

∴ P (a blue marble and a king) = P(a blue marble) + P(a king) = $$\frac 35$$ + $$\frac 1{13}$$ = $$\frac {39 + 5}{65}$$ = $$\frac {44}{65}$$Ans

Here,

Total number of cards = 52

1. In the first time,
P(a diamond) = $$\frac {13}{52}$$ = $$\frac 14$$
In the second time,
P(a diamond) = $$\frac {13}{52}$$ = $$\frac 14$$
∴ P(diamond both times) = P(a diamond)× P(a diamond) = $$\frac 14$$× $$\frac 14$$ = $$\frac 1{16}$$Ans
2. In the first time,
P(a diamond) = $$\frac {\text{number of diamond in the pack}}{\text{total number of cards}}$$ = $$\frac {13}{52}$$ = $$\frac 14$$
In the second time,
P(a heart) = $$\frac {\text{number of heart in the pack}}{\text{total number of cards}}$$ = $$\frac {13}{52}$$ = $$\frac 14$$
∴ P(a diamond and a heart) = P(a diamond)× P(a heart) = $$\frac 14$$× $$\frac 14$$ = $$\frac 1{16}$$Ans

When a coin is tossed, the sample space S has two sample points, S = {H, T} where H and T denote Head and Tail respectively.

∴ n(S) = 2

Now,

P(T) = $$\frac {n(T)}{n(S)}$$ = $$\frac 12$$

Since, in a single throw of a dice, the sample space S has six sample points,

S = {1, 2, 3, 4, 5, 6}

∴ n(S) = 6

Here,

P(4) = $$\frac {\text{number of face 4 in S}}{n(S)}$$ = $$\frac 16$$

Therefore,

P(T and 4) = P(T)× P(4) = $$\frac 12$$× $$\frac 16$$ = $$\frac 1{12}$$Ans

0%

(frac 14)

(frac 12)

(frac 84)

1

1

(frac 45)

(frac 25)

(frac 75)

(frac 59)

(frac 79)

1

(frac 29)

(frac 7{10})

(frac 2{10})

(frac 3{10})

1

• ### A card is drawn randomly from a pack of cards and a dice is thrown once. Determine the probability of not getting a card of king as well as 6 on the dice.

(frac {1}{13})

(frac{10}{13})

(frac {13}{10})

1

1

(frac {12}1)

;i:4;s:22:

(frac 1{12})

• ### The king, queen and jack are removed from a deck of 52 playing cards and then well suffled. One card is selected from the remaining cards. Find the probability of getting the numbered 10 or ace.

(frac {49}{52})

(frac {3}{49})

(frac 51)

(frac 15)

(frac 32)

1

(frac 15)

(frac 23)

• ### A deck of cards in which a card of diamond numbered 4 is replaced instead of a card of club numbered 5. What is the probability of getting the numbered 4 or face cards.

(frac {17}{52})

(frac {52}{17})

(frac {4}{17})

1

1

17

8

5

(frac 16)

(frac 35)

(frac 56)

1

(frac 5{26})

(frac 1{26})

(frac 7{26})

1

1

(frac 15)

(frac3{50})

(frac 1{10})

• ### There are 40 students in a class. Among them, 15 are boys and remaining girls. If 15 of them study Optional Maths, find the probability of selecting a girl studying Optional Maths.

1

(frac {15}{64})

(frac {64}{15})

(frac {15}{35})

• ### In a survey of student, it is found that 60% ride cycle and 30% do not play CD player. Find the probability of selecting a student riding cycle but playing the CD player.

(frac {21}{50})

(frac {50}{21})

(frac {30}{50})

1

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