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As 3 dice are thrown. So total possible outcomes = 6×6×6 = 216 

For getting 15 as the sum we need this combination- 

(5, 5, 5) or (6,4,5) or (6,6,3) 

(6, 4, 5) can be arranged in 3! = 6 ways. 

(6, 4, 5) can be arranged in (3!)/(2!) = 3 ways 

∴ 10 favourable outcomes are possible. 

∴ P(E) =  \(\frac{10}{216}\)

Correct option is: A) 0

Impossible event is an event which can not happen.

Also, the probability of an impossible event is 0.

Correct option is: A) 0

A year which is not a leap year is Normal year having 365 days.
There are 52 weeks, i.e., 52 × 7 = 364 days and 1 day extra in a normal year.

So, the sample space for 1 extra day is expressed as follows:

U = {Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday}

∴ Total primary outcomes n = 7

A = Event that having 53 Fridays in a year which is not a leap year.= {Friday}

∴ Favourable outcome for the event A is m = 1

Hence, P(A) = \(\frac{m}{n} = \frac{1}{7}\)

Correct option is: C) sample space

The set of all possible event is called the sample space.

Correct option is: C) sample space

The correct option is: (C) 2/7

Explanation:

 Leap year contains 366 days. 

=> 52 weeks + 2 days 

52 weeks contain 52 Fridays. 

We will get 53 Fridays if one of the remaining two days is a Friday. Total possibilities for two days are: 

(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday) 

There are 7 possibilities and out of these 2 are favourable cases

∴ Required probability = 2/7

Correct option is: B) \(\frac 27\)

2016 is a leap year in which total number of days is 366.

\(\because\) \(\frac {366}7 = 52 + \frac 27\) 

\(\therefore\) Year 2016 definitely have 52 Mondays 

and there are two extra days left which may be 

(i) Monday & Tuesday, (ii) Tuesday & Wednesday (iii) Wednesday & Thursday (iv) Thursday & Friday (v) Friday & Saturday (vi) Saturday & Sunday (vii) Sunday & Monday.

For 53 Mondays there are two possibilities, left 2 days  may be Sunday & Monday or Monday & Tuesday.

\(\therefore\) Total favourable outcomes which favours 53 Mondays = 2

Total possible outcomes = 7.

\(\therefore\) Probability that year 2016 to have 53 Mondays 

= \(\frac {Total \, favourable \, outcome}{Total \, No.\, of \,outcomes}\) = \(\frac 27\) 

Hence, the probability that year 2016 to have 53 Mondays is \(\frac 27\)

Correct option is: B) \(\frac{2}{7}\)

Correct option is: B) \(\frac 27\)

Number of days in a leap year = 366.

\(\because\) \(\frac {366}7 = 52 + \frac 27 = \) 52 weeks and 2 days remaining.

Hence, in a leap year there is definitely 52 Fridays but 53 Fridays depends on 2 remaining days. 

There are total 7 outcomes for these two remaining days.

They can be (i) Sunday & Monday, (ii) Monday & Tuesday (iii) Tuesday & Wednesday

(iv) Wednesday & Thursday (v) Thursday & Friday (vi) Friday & Saturday

(vii) Saturday & Sunday

But favourable outcomes for 53 Fridays are Thursday & Friday or Friday & Saturday.
Hence, there are 2 favourable outcomes for 53 Fridays out of 7 total outcomes.

\(\therefore\) Probability of getting 53 Fridays in a leap year is 

P = \(\frac {Total \, favourable \, outcome}{Total\, possible \, outcomes}\) = \(\frac 27\)

Correct option is: B) \(\frac{2}{7}\)

As 2 dice are thrown so there are 6×6 = 36 possibilities. 

Let E denote the event of getting a total score of 5. 

E = {(1,4),(2,3),(3,2),(4,1)} 

∴ n(E)=4 

Hence, 

P(E) = \(\frac{4}{36} = \frac{1}{9}\)

As our answer matches only with option (c) 

∴ Option (c) is the only correct choice.

When a coin is tossed three times, all possible outcomes are 

HHH,HHT,HTH,THH,HTT,THT,TTH and TTT.

The number of total outcomes = 8.

(i) The outcomes with three heads is HHH.

the number of outcomes with three heads = 1

Therefore, P(getting at least two tails) = \(\frac{number\, of\,favorable\,outcomes}{number\,of \,all\,possible\,outcomes}\) = \(\frac{4}5\) = \(\frac{1}2\)

Thus, the probability of getting at least two tails is \(\frac{1}2\).

Therefore are 6 red face cards. These are removed.

Thus, remaining number of card = 52 - 6 = 46.

(i) number of red cards  now = 26 - 6 = 20.

Therefore, P(getting a red card) = \(\frac{number\, of\,favorable\,outcomes}{number\,of \,all\,possible\,outcomes}\) = \(\frac{20}{46}\) = \(\frac{10}{23}\)

Thus, the probability  that the drawn card is a red card is \(\frac{10}{23}\).

(ii) Number of face cards now = 12 - 6 = 6.

Therefore, P(getting a face card) = \(\frac{number\, of\,favorable\,outcomes}{number\,of \,all\,possible\,outcomes}\) = \(\frac{6}{46}\) = \(\frac{3}{23}\).

Thus, the probability  that the drawn card is a face card is \(\frac{3}{23}\).

(iii) The number of cards of clubs  = 12.

Therefore, P(getting a card of clubs) = \(\frac{number\, of\,favorable\,outcomes}{number\,of \,all\,possible\,outcomes}\) = \(\frac{12}{46}\) = \(\frac{6}{23}\)

Thus, the probability  that the drawn card is a card of clubs is \(\frac{6}{23}\).

It is the performance of an experiment, such as throwing a dice or tossing a coin.

It is the chance of happening of an event when measured quantitatively.

Given that, the probability that A will die = p

The probability that B will die = q

The probability that only one of them will be alive = P(A∩B’) + P(A’∩B)

As, the events are independent, the above form can be written as:

P(A∩B’) + P(A’∩B) = P(A).P(B’) + P(A’)P(B)

P(A∩B’) + P(A’∩B) = p (1-q) + (1-p)(q)

P(A∩B’) + P(A’∩B) = p+q-2pq.

Let E denote the event that the chosen numbers are consecutive. 

No of ways in which 3 numbers can be chosen out of 30 = ³⁰C₃ 

As we have to select 3 consecutive numbers, if we select 1 number other two are already selected. 

As 29,30 can’t be selected because if they are selected we won’t be able to get 3 consecutive numbers. 

∴ number of ways in which 3 consecutive numbers can be selected = number of ways in which 1 number can be chosen out of numbers from 1 to 28 = ²⁸C₁ ways 

∴ P(E) = \(\frac{28}{^{30}C_3}\) = \(\frac{28\times3\times2\times1}{30\times29\times28}\) = \(\frac{1}{145}\)

Thus, P(E) =   \(\frac{1}{145}\)

Let E denote the event that the selected persons are sitting together.

As 2 persons can be selected out of n in ⁿC₂ ways 

Out of n persons we can select two persons sitting together in (n-1) ways. 

Because we have to select only one person next person is going to be automatically selected. 

We can’t select last person because no one is sitting next to him. 

∴ 1 person out of n-1 persons can be selected in (n-1) ways. 

∴ P(E) = \(\frac{n-1}{^nC_2}\)  = \(\frac{2(n-1)}{n(n-1)}\) = \(\frac{2}{n}\)

Thus, P(E) = \(\frac{2}{n}\) 

Conditional Probability: Let A and B be two events associated with a random experiment. Then, the probability of the occurrence of A under the condition that B has already occurred and P(B) ≠ 0, is called the conditional probability of A given B and is written as P(A/B).

Example:

Suppose a red card is drawn from a pack of 52 cards, and is not put back, then the probability of drawing a red card in the first attempt is \(\frac{26}{52}\) and in the second one it is \(\frac{25}{51}\) as the red card is not replaced. 

Similarly in the above given case, if we draw a black card in the second attempt, then its probability = \(\frac{26}{51}\) as number of black cards = 26 but total number of remaining cards = 51.

 Hence the occurrence of the second event is fully dependent on the first event. 

Such events are called conditional events.

If the event A occurs when B has already occurred, then P(B) ≠ 0, then we may regard B as a new (reduced) sample space for event A. 

In that case, the outcomes favourable to the occurrence of event A are those outcomes which are favourable to B as well as favourable to A, i.e, the outcomes favourable to A ∩ B and probability of occurrence of A so obtained is the conditional probability of A under the condition that B has already occurred. 

∴ P(A/B) = \(\frac{Number\,of\,outcomes\,favourable\,to\,both\,A\,and\,B}{Number\,of\,outcomes\,in\,sample\,space(B,here)}\)

=\(\frac{n(A\,\cap\,B)}{n(B)}\) =   \(\frac{\frac{n(A\,\cap\,B)}{n(S)}}{\frac{n(B)}{n(S)}} \) = \(\frac{P(A\,\cap\,B)}{P(B)}\) ,  where S is the sample space for the events A and B.

Similarly, 

P(B/A) = \(\frac{P(A\,\cap\,B)}{P(A)}\) , P(A) ≠ 0 

where P(B/A) is the conditional probability of occurrence of B, knowing that A has already occurred. 

Note: If A and B are mutually exclusive events, then,

 P(A/B) = \(\frac{P(A\,\cap\,B)}{P(B)}\) = 0 ∵ \(P(A\,\cap\,B)\) = 0

P(B/A) =\(\frac{P(A\,\cap\,B)}{P(A)}\) =0 ∵ \(P(A\,\cap\,B)\) =0

(B)  A coin is tossed successively three times. The probability of getting one head or two heads is 3/4.

(i) Random experiment: An experiment is called random experiment if it has more than one possible outcome and it is not possible to predicts the outcome in advance.

(ii) Outcomes: A possible result of a random experiment is called its outcome. 

(iii) Sample space: The set of all possible outcomes of a random experiment is called the sample space associated with the experiment. It is denoted by the symbol S. 

(iv) Sample point: Each element of a sample space is called sample point, or each outcome of the random experiment is called sample point. 

Correct option is: D) sure event

When a dice is rolled the possible outcomes are {1, 2, 3, 4, 5, 6}

Total No of outcomes is n (S) = 6

Total No of numbers less than or equal to 6 = n(E) = 6

\(\therefore\) Probability of getting a number less than or equal to 6 is 

P = \(\frac {n(E)}{n(S)} = \frac 66 = 1\) which is probability of sure event.

\(\therefore\) The event of getting a number less than or equal to 6 in a dice rolling is a sure event.

Correct option is: D) sure event

Correct option is: A) Event

Collection of outcomes of an experiment is event.

Correct option is: A) Event

An operation which can produce some well defined outcomes in known as an experiment.

Any experiment that results in two to more outcomes is called a random experiment.

The set of all possible outcomes of a random experiment is called as sample space & it is denoted by S.

The probability of sample space is P(S) = 1.

Any subset of a sample space is called an event.

Range of probability is 0 ≤ P(A) ≤ 1.

p(certain event) = 1

P(Impossible event) = 0

(a) If two events cannot occur simultaneously in a random experiment then they are called mutually exclusive events, 

(b) Events are said to be independent, if the occurrence of one does not depend upon the occurrence of the other, 

(c) In a random experiment, let S be the sample space & let E be an event then E C S, So ‘E’ is also an event called the complementary of E.