# Probability Chapter 15 Extra Questions and Solutions For Class 10 CBSE Mathematics

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## Solution.

Probability that it will be foggy tomorrow P(E) = 0.85. Required to find, probability that it will not be foggy tomorrow P(E’). We know that sum of the probability of occurrence of an event and the probability of non-occurrence of an event is 1.

So,

P(E) + P(E’) = 1

0.85 + P(E’) = 1

P(E’) = 1 – 0.85

P(E’) = 0.15

Therefore, the probability that it will not be foggy tomorrow is =  0.15

## Solution

Three coins are tossed simultaneously. When three coins are tossed then the outcome will be anyone of these combinations.So, the possible outcomes are:

TTT, THT, TTH, THH. HTT, HHT, HTH, HHH.

So, the total number of outcomes is 8.

(i) For exactly two heads, the favourable outcome are THH, HHT, HTH

So, the total number of favourable outcomes is 3. We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting exactly two heads is  3/8.

(ii) For getting at least two heads the favourable outcomes are HHT, HTH, HHH, and THH

So, the total number of favourable outcomes is 4. We know that, Probability = Number of favourable outcomes/ Total number of outcomes.

Thus, the probability of getting at least two heads when three coins are tossed simultaneously =  4/8 =  1/2.

## Solution.

Given A pair of dice is thrown. We have to find the probability that the total of numbers on the dice is greater than 10. First, let’s write the all possible events that can occur:

(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6),

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6),

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6).

It’s seen that the total number of events is 62 = 36 . Favourable events i.e. getting the total of numbers on the dice greater than 10 are (5, 6), (6, 5) and (6, 6). So, the total number of favourable events i.e. getting the total of numbers on the dice greater than 10 is 3.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes. Thus, the probability of getting the total of numbers on the dice greater than 10  =  3/36  =  1/12.

## Solution.

A card is drawn at random from a pack of 52 cards. We have to find the probability of the following….

Total number of cards in a pack = 52.

1. the probability of getting a black king = 2/52 = 1/26.
2. A jack, queen or a king are 3 from each 4 suits.

So, the total number of a jack, queen and king are 12.

We know that, Probability = Number of favorable outcomes/ Total number of outcomes

Thus, the probability of getting a jack, queen or a king is 12/52 = 3/13

3. Total number of black cards in the pack is 26

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting black cards is 26/52 = 1/2.

4. Total number of red cards is 26

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a red card = 26/52 = 1/2

## Solution.

It is given that an urn contains 10 red and 8 white balls. It is required to find probability that one ball is drawn at random and getting a white ball.

Total number of balls 10 + 8 = 18. Total number of white balls is 8.

We know that, Probability = Number of favorable outcomes/ Total number of outcomes.

Therefore, the probability of drawing a white ball from the urn is  8/18 =  4/9.

## Solution.

It is given that the numbers are from 1 to 15. One number is selected randomly.

We need to find the probability that the selected number is a multiple of 4

Total number between from 1 to 15 to 15. Numbers that are multiple of 4 are 4, 8 and 12. We know that, Probability = Number of favourable outcomes/ Total number of outcomes.

Thus, the probability of selecting a number which a multiple of 4 is 3/15 = 1/5.

## Solution:

It is given that in a lottery there are 10 prizes and 25 blanks. We need to find probability of winning a prize

Total number of tickets is 10 + 25 = 35

Total number of prize carrying tickets is 10

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of winning a prize  =  10/35  =  2/7.

## Solution.

1. Favorable outcomes i.e. to get an odd number are 1, 3, 5, 7, 9, and 11

So, total number of favourable outcomes i.e. to get a prime number is 6

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a prime number = 6/12 = 1/2  .

2.  Favourable outcomes i.e. to get an even number are 2, 4, 6, 8, 10, and 12

So, total number of favourable outcomes i.e. to get an even number is 6

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting an even number = 6/12 = 1/2  .

## Solution:

No. of possible outcomes while tossing a coin = 2 i.e., 1 head or 1 tail

Probability = Number of favourable outcomes / Total number of outcomes,

P (getting tail) = 1/2

As we can see that the probability of both the events are equal, these are called equally like events.

Thus, tossing a coin is considered to be a fair way of deciding which team should choose ends in a game of cricket.

## Solution.

Given numbers are 1, 2, 2, 3, 3, 3, 4, 4, 4, 4 . Total number of possible outcomes = 10

Average of the numbers = ( 1+ 2 + 3 + 3 + 3 + 4 + 4 + 4 + 4 ) / 10 = 30/10 = 3.

Now, let E be the event of getting 3.

Number of favourable outcomes = 3 {3, 3, 3}

P(E) = Number of favourable outcomes/ Total number of outcomes

P(E) = 3/10

Therefore, the probability that a number selected at random will be the average is 3/10.

## Solution.

Total number of balls  =  8 + 6 + 4 = 18. Total no. of possible outcomes  = 18.

i) Let E = Event of getting red or white ball

No. of favourable outcomes = 14 ( 8 red balls +  6 white balls)

Probability, P(E) = Number of favourable outcomes/ Total number of outcomes

P(E) = 14/18

P(E) = 7/9.

ii) Let E = event of getting neither a white nor a black ball

No. of favourable outcomes = 18 – 6 – 4

= 8 (Total balls –  no. of white balls –  no. of black balls)

Probability, P(E) = Number of favourable outcomes/ Total number of outcomes

P(E) = 8/18 =  4/9.

## Solution.

Numbers from 1, 2, 3….. 35 are a total of 35. Total no. of possible outcomes = 35.

(i) Let E = event of getting a prime number

No. of favorable outcomes = 11  {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31)

Probability, P(E) = Number of favourable outcomes/ Total number of outcomes

P(E) = 11/35.

(ii) Let E = event of getting a number which is a multiple of 7

No. of favourable outcomes = 5 {7, 14, 21, 28, 35}

P(E) = Number of favourable outcomes/ Total number of outcomes

P(E) = 5/35 = 1/7.

## Solution.

No. of good pens = 144 – 20 = 124. No. of detective pens = 20 . Total no. of possible outcomes =144 (total no. of pens).

So, for her to buy it the pen should be a good one. let E = event of buying a pen which is good.

No. of favourable outcomes = 124 (124 good pens)

P(E) = Number of favourable outcomes/ Total number of outcomes

P(E) = 124/144 = 31/36 .

So, the probability that Nomii will buy the pen is 31/36 .

## Solution.

We have,

No. of good pens = 132. No. of defective pens = 12 . So, the total no. of pens = 132 + 12 = 144 .

Then, the total no. of possible outcomes = 144.

Now, let E = event of getting a good pen.

No. of favorable out comes = 132 {132 good pens}

P(E) = Number of favorable outcomes /  Total number of outcomes

P(E) = 132/144 = 11/12.

## Solution.

(i) an orange flavored candy

We know that the bag contains lemon flavored candies only. So, the event that Mahi will take out an orange flavored candy is an impossible event.

Thus, the probability of impossible event is 0 .

P(an orange flavored candy) = 0.

(ii) a lemon flavored candy

As the bag contains lemon flavored candies only. Then, the event that Mahi will take out a lemon flavoured candy is sure event. Thus, the probability of sure event is 1.

P(a lemon flavoured candy) = 1.

## Solution.

A bag contains 3 red and 5 black balls.

So, the total no. of possible outcomes = 8 (3 red + 5 black).

Let E = event of getting red ball.

No. of favourable outcomes = 3 (as there are 3 red)

P(E) = Number of favorable outcomes/ Total number of outcomes

P(E) = 3/8.

Now,

P(E) + P(E’) = 1

P(E’) = 1 – P(E)

P(E’)= 1 –  3/8

P(E’) = 5/8 .

So, 5/8  is the probability of not getting red balls.

## Solution.

A box containing 5 red, 8 white and 4 green marbles. So, the total no. of possible outcomes = 17 (5 red + 8 white + 4 green).

Let E= event of getting a green marble

Number of favourable outcomes = 4 (as 4 green marbles)

Probability, P(E) = Number of favourable outcomes/ Total number of outcomes

P(E) = 4/17

So,

P(E) + P(E’) = 1

P(E’) = 1 – P(E)

P(E’)= 1 –  4/17

P(E’) = 13/17 .

Therefore, the probability that the marble taken out is not green is 13/17.

## Solution.

Area of a circle with radius 0.5 m A circle = (0.5)2 = 0.25 πm2 . Area of rectangle = 3 x 2 = 6m2.

Probability ( Geometric ) = ( measured specific region ) / ( measured whole region).

Probability ( Geometric ) =  ( area of the circle in violet )/ ( area of the rectangle in olive)

=   0.25 πm2 /  6m2.

=    π / 24 .

Therefore, the probability that the tie will land inside the circle = π/24.

## Solution.

In a class there are 18 girls and 16 boys, the class teacher wants to choose one name. The class teacher writes all pupils’ name on a card and puts them in basket and mixes well thoroughly. A child picks one card. It is required to find the probability that the name written on the card is a girl’s name …….

Total number of students in the class = 18 + 16 = 34.

The names of a girl are 18, so the number of favourable cases is 18

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a name of girl on the card = 18/34 = 9/17.

## Solution.

It is given that one card is drawn from a well shuffled deck of 52 playing cards.We have to find the  probability of a king in red suit. Total number of cards is 52

Total number of cards which are king of red suit is 2

Number of favourable outcomes i.e. Total number of cards which are king of red suit is 2.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting cards which is a king of red suit = 2/52 = 1/26.