Chapter-15, Probability

Explanation:

(i) Probability of event E + Probability of event “not E” = 1

(ii) The probability of an event that cannot happen is 0. Such an event is called an impossible event.

(iii) The probability of an event that is certain to happen is 1. Such an event is called a sure or a certain event.

(iv) The sum of the probabilities of all the elementary events of an experiment is 1.

(v) The probability of an event is greater than or equal to 0 and less than or equal to 1.

Explanation:

(i) "A Driver Attempts to Start a Car" is the experiment's first scenario. We are not justified in assuming that each possible event is equally likely to occur when the car starts or does not start. Hence, no two results of the experiment are equally plausible.

(ii) In the experiment, a basketball is shot by a player. We are not justified in assuming that each possible outcome is equally likely to occur when we say, "She/he shoots or misses the shot." Hence, no two results of the experiment are equally plausible.

(iii) During the test To respond to a true-false question, a trial is created. There is a proper or wrong response. We are aware in advance that the outcome may go one of two ways: either in the correct direction or the wrong direction. It is reasonable to presume that both good and bad outcomes are equally likely to occur. Right or wrong outcomes are therefore equally likely.

(iv) "A Baby is Born, It Is a Boy or a Girl" is the experiment. We are aware that the result could result in either a boy or a girl, one of the two potential results. It is reasonable for us to believe that both the boy and the girl outcomes are equally likely to occur. Boy or girl outcomes are therefore equally likely.

Explanation:

Given that there are only two possible outcomes for a coin toss, either head up or tail up, it is thought to be a fair technique to determine which team should receive the ball to start a football game. It is reasonable to infer that each outcome—head or tail—has the same chance of happening as the other, i.e., both events are equally likely. Hence, the outcome of tossing a coin is totally random.

Explanation:

(B) Because the probability of an event E is represented by the number P(E),

0 P(E) 1

The probability of an event cannot be

Explanation:

P(E)+P(not E) = 1

As presented, P(E) = 0.05

So, P(not E) = 1-P(E)

Or, P(not E) = 1-0.05

∴ P(not E) = 0.95

Explanation:

(i) Take into account the incident with the experiment of removing an orange-flavored candy from a bag that solely contained lemon-flavored candies. It is impossible for there to be an outcome that produces orange-flavored candy, hence its probability is zero.

(ii) Imagine removing a lemon-flavored candy from a bag that solely contained lemon-flavored candies. This event's probability is 1, as it is a certain event.

Explanation:

A similar birthday occasion is represented by E.

P(E) = 0.992

But P(E) + P = 1

P = 1 – P(E) = 1 – 0.992 = 0.008.

Explanation:

A bag has 3 + 5 = 8 balls. There are eight different methods to choose one of these balls.

Eight elementary events total.

(i) There are three methods to draw one red ball because there are three of them in the bag.

3 is the ideal number of elementary events.

P (obtaining a red ball) so equals

(ii) One black ball (not a red ball) can be drawn in five different ways because the bag contains five black balls and three red balls.

5 is the ideal number of elementary events.

P (obtaining a black ball) so equals 

Explanation:

There are a total of 17 marbles in the box (5 + 8 + 4).

There were 17 elementary events in total.

(i) The box contains five red marbles.

5 is the ideal number of elementary events.

Getting a red marble P equals .

(ii) The package contains 8 white marbles.

8 is the ideal number of elementary events.

Getting a white marble P equals

(iii) The box contains 5 + 8 = 13, none-green marbles.

13 is the ideal number of elementary events.

P (failure to receive a green marble) =

Explanation:

180 coins total in a piggy bank = 100 + 50 + 20 + 10

180 total elementary events were recorded.

(i) The piggy bank has 150 coins in total.

100 is the ideal number of elementary events.

P (dropping off a 50 pence coin) = =

(ii) In addition to the Rs. 5 coins, there are 100 + 50 + 20 = 170 coins.

170 is the ideal number of elementary events.

P(A coin other than a Rs. 5 coin dropping out) = = 

Explanation:

There are a total of 13 fish in the aquarium (5 + 8).

There were 13 elementary events in total.

The tank has 5 male fish.

5 is the ideal number of elementary events.

Hence, P (removing a male fish) =.

Explanation:

An arrow can point any of the 8 integers in any of the 8 possible directions.

Eight favourable outcomes in total.

(i) One outcome was favourable.

P (arrow points at 8) =

(ii) Four outcomes were favourable.

P (arrow points at an odd number) =

(iii) The number of favourable outcomes is 6.

P (arrow points to a number > 2) =

(iv) 8 outcomes were favourable.

P (arrow points at a number< 9) = =1.

Explanation:

Six outcomes are favourable overall when rolling the dice.

(i) The prime numbers on a die are 2, 3, and 5.

Hence, favourable outcomes equal 3.

P thus equals acquiring a prime number

(ii) The numbers on a dice between 2 and 6 are 3, 4, and 5.

Hence, favourable outcomes equal 3.

Hence, P (obtaining a number between 2 and 6) =

(iii) The odd numbers on a die are 1, 3, and 5.

Hence, favourable outcomes equal 3.

P thus equals an odd number

Explanation:

There were 52 positive results in all.

(i) Red cards come in two suits: diamond and heart. One king is present in each suit.

Favourable results equal one.

P (a red-coloured king) =

(ii) Each pack contains 12 face cards.

Favourable results equal 12

P (a face card) hence equals

(iii) The two suits of red cards are heart and diamond. There are 3 face cards in each suit.

Positive results =  6

P (a red face card) so equals

(iv) There is only one jack of hearts

Favourable result = 1

P (the jack of hearts) thus equals

(v) There are 13 spade cards in total.

Favourable results equal 13

P (a spade) hence Equals

(vi) The queen of diamonds is a singular being.

Favourable result = 1

P (the queen of diamonds) hence equals

Explanation:

Five favourable outcomes in total.

(i) There is just one queen.

Favourable result = 1

P (queen) =

(ii) In this case, there are a total of 4 favourable outcomes.

(a) A positive result equals 1.

P (an ace) Equals

(b) No card qualifies as the queen.

Favourable result = 0

P (queen) =.

Explanation:

132 + 12 = 144 is the total number of favorable outcomes.

132 favorable outcomes in total.

Hence, P (acquiring a quality pen) =.

Explanation:

(i) There were 20 positive results in all.

There are 4 favorable outcomes in total.

Hence, P (obtaining a faulty bulb) =

(ii) The total number of positive results is now 20 - 1 = 19.

Number of positive results = 19 - 4 = 15

P (obtaining a non-faulty bulb) hence

Explanation:

There were 90 positive results in all.

(i) 90 - 9 = 81 is the number of two-digit numbers from 1 to 90.

Favorable results equal 81

P thus receiving a disc with a two-digit number =

(ii) The perfect squares from 1 to 90 are 1, 4, 9, 16, 25, 36, 49, 64, and 81.

Positive results = 9

P (obtaining a perfect square) =

(iii) There are 18 numbers from 1 to 90 that are divisible by 5.

Favorable results equal 18

P (obtaining a value that may be divided by 5) =

Explanation:

There were six positive results in total.

(i) There were two positive results.

Hence, P(Receiving a Letter A)=

(ii) The number of positive results is 1.

Thus P (After receiving the letter D)= 

Explanation:

The overall area of the rectangle-shaped supplied figure =

circle's area = = = m2

P (die to land inside the circle) hence equals

Explanation:

There were 144 successful results in all.

(i) A Total number of Nondefective pens is 124 (144 - 20).

124 are the favorable outcomes in total.

As a result, P (she will purchase) = P (a pen without a flaw) =

(ii) There were 20 favorable outcomes.

Thus, P (she won't purchase) = P (a bad pen) =

Explanation:

Two dice are thrown, and the total favorable outcomes are:

(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)

There were 36 positive results in all.

(i) The benefits of obtaining the sum as 3 = 2 lead to P (obtaining the amount as 3) =

Positive results of obtaining the amount as 4 = 3 Hence, P (obtaining the sum as 4) =

Benefits of determining that the sum is 5 = 4

P  (obtaining the sum as 5)

Positive results of obtaining the amount as 6 = 5 Hence, P (obtaining the sum as 6) =

Benefits of calculating the amount as 7 = 6

P(obtaining the sum as 7)

Positive results of obtaining the amount as 9 = 4 Hence, P (obtaining the sum as 9) =

Positive results of obtaining the amount as 10=3; hence, P (obtaining the sum as 10) =

Positive results of obtaining the amount as 11 = 2; hence, P (obtaining the sum as 11) =

(ii) I disagree with the justification offered here. Some of the justification has already been provided.

Explanation:

Results of the experiment in which a coin is tossed three times:

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

Hence, there were 8 positive results overall.

6 favorable outcomes were achieved.

Hence, needed probability =.

Explanation:

(i) The experiment in which a die is thrown twice has the following results:

(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)

As a result, 36 favorable results total.

Now think about the following occurrences:

The first toss indicates 5 in A, and the second throw indicates 5 in B.

Hence, there are six favorable possibilities in each scenario.

P(A) = and P(B) =

Pand P

Needed probability =

(ii) Let S represent the sample area connected to the two-die rolling experiment. Thus n(S) = 36.

AB stands for getting five in each throw, or five in the first and second throws.

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

And B = (1, 5) (2, 5) (3, 5) (4, 5) (5, 5) (6, 5)

P(A) = , P(B) = and P(AB) =

Needed probability is the likelihood that at least one of the two throws will result in a five.

= P (AB) = P(A) + P(B) – P(AB)

= 

Explanation:

(i) False: We can categorise the outcomes in this way, but they are not then "equally likely." Because "one of each" can come from either a head and a tail on the first coin or a tail and a head on the second coin, there are two possible outcomes. This increases the likelihood of two heads by two (or two tails).

(ii) True: The probabilities of the two outcomes taken into account in the question are equal.

Explanation:

The overall positive results of the random experiment in which two consumers, Shyam and Ekta, visited a certain shop from Tuesday through Saturday are as follows:

(T, T) (T, W) (T, TH) (T, F) (T, S)

(W, T) (W, W) (W, TH) (W, F) (W, S)

(TH, T) (TH, W) (TH, TH) (TH, F) (TH, S)

(F, T) (F, W) (F, TH) (F, F) (F, S)

(S, T) (S, W) (S, TH) (S, F) (S< S)

There were 25 positive results in all.

(i) Visiting on the same day has the following positive effects: (T, T), (W, W), (TH, TH), (F, F), and (S, S).

5 is the number of successful outcomes.

Hence, needed probability =.

(ii) Visiting on consecutive days has favorable effects for (T, W), (W, T), (W, TH), (TH, W), (TH, F), (F, TH), (S, F), and (F, S).

8 favorable outcomes in total.

Hence, needed probability =

(iii) The number of positive results from visiting on various days is 25 - 5 = 20.

20 favorable outcomes in total.

Hence, needed probability =.

Explanation:

The whole table is as follows:

It is evident that 36 is the total number of positive outcomes.

(i) There are 18 favourable consequences for achieving an equal score overall.

Hence, P (achieving an even score overall) =

(ii) There are four favourable outcomes from receiving a total score of six.

P (obtaining a total score of 6)=

(iii) There are 15 favourable outcomes for achieving a total score of at least 6.

P (obtaining a total score of at least 6) =

Explanation:

Allow x blue balls to be in the bag.

Number of balls in the bag overall =

= Drawing a blue ball’s Probability =

= Drawing a blue ball’s Probability =

=

=2

Thus, the bag contains 10 blue balls.

Explanation:

The box contains 12 balls.

Hence, there were 12 favourable results overall.

The proportion of successful results is 

Hence, P (obtaining a black ball) =

If there are added 6 balls to the box, then

An overall number of positive results: 12 + 6 = 18.

And the number of favourable results =

Getting a black ball equals P2 = P=

According to the enquiry,

=

=2

Explanation:

Total number of favorable outcomes in this case = 24

Green marbles are to be present =

Hence, favorable outcomes=

P(G) =

Yet P(G) =

= 16

Hence, there are 16 green marbles.

The amount of blue marbles is equal to 24 - 16 = 8.