1 Marks Question - Page 2 - Maths STD 11 Science Questions - Vidyadip

Questions · Page 2 of 2

1 Marks Question

Question 511 Mark

Three coins are tossed. Describe Three events which are mutually exclusive and exhaustive.

Answer

When three coins are tossed then sample space (S) is given by

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Now, let A be the event: getting at least two heads = {HHH, HHT, HTH, THH}
B be the event: getting exactly one head = {HTT, THT, TTH}
and C be th event: getting no head = {TTT}

Mutually exclusive events are those in which no element is common
Since ${A\cap B=\phi}$. $\therefore$ A and B are mutually exclusive events,
$\style{font-size:28px}{B\cap C=\phi}$. $\therefore$ B and C are mutually exclusive events,
$\style{font-size:28px}{A\cap C=\phi}$. $\therefore$ A and C are mutually exclusive events.
Thus A, B, C are mutually exclusive events.
Also, as ${A\cup B\cup C=\{HHH,\;HHT,\;HTH,\;THH,\;HTT,\;THT,\;TTH,\;TTT\}=S}$
So, A, B, C are exhaustive events.
A, B, C are three events which are mutually exclusive and exhaustive.

View full question & answer→

Question 521 Mark

Three coins are tossed. Describe Two events which are mutually exclusive.

Answer

When three coins are tossed then sample space (S) is given by S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Let event A: getting at least two heads = {HHH, HHT, HTH, THH} and
event B: getting at least two tails = {HTT, THT, TTH, TTT}
There should not be any element common for the events to be mutually exclusive.
since ${A\cap B=\phi}$
Thus A and B are mutually exclusive events.

View full question & answer→

Question 531 Mark

A die is rolled. Let E be the event die shows 4 and F be the event die shows even number. Are E and F mutually exclusive?

Answer

When a die is rolled then S = {1, 2, 3, 4, 5, 6}
E: die shows 4 = {4}
F: die shows even number = {2, 4.6}
Now $E \cap F = (4) \ne \phi$
Thus E and F are not mutually exclusive events.

View full question & answer→

Question 541 Mark

Let a sample space be S = {$\omega_1, \omega_2,..., \omega_6$}. Which of the following assignments of probabilities to each outcome are valid?

Outcomes	$\omega_1$	$\omega_2$	$\omega_3$	$\omega_4$	$\omega_5$	$\omega_6$
	0.1	0.2	0.3	0.4	0.5	0.6

Answer

Since, sum of probabilities = 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 = 2.1.
Hence, the assignment is not valid.

View full question & answer→

Question 551 Mark

Let a sample space be S = {$\omega_1, \omega_2,..., \omega_6$}.Which of the following assignments of probabilities to each outcome are valid?

Outcomes	$\omega_1$	$\omega_2$	$\omega_3$	$\omega_4$	$\omega_5$	$\omega_6$
	$\frac 1{12}$	$\frac 1{12}$	$\frac 16$	$\frac 16$	$\frac 16$	$\frac 32$

Answer

We know that in no case P > 1.
Since p($\omega_6$) = $\frac32$> 1, the assignment is not valid.

View full question & answer→

Question 561 Mark

Let a sample space be S = {$\omega_1, \omega_2,..., \omega_6$}.Which of the following assignments of probabilities to each outcome are valid?

Outcomes	$\omega_1$	$\omega_2$	$\omega_3$	$\omega_4$	$\omega_5$	$\omega_6$
	$\frac 18$	$\frac 23$	$\frac 13$	$\frac 13$	$-\frac 14$	$-\frac 13$

Answer

Clearly we see that two of the probabilities p($\omega_5$) and p($\omega_6$) are negative so the assignment is not valid.

View full question & answer→

Question 571 Mark

Let a sample space be S = {$\omega_1, \omega_2,..., \omega_6$}.Which of the following assignments of probabilities to each outcome are valid?

Outcomes	$\omega_1$	$\omega_2$	$\omega_3$	$\omega_4$	$\omega_5$	$\omega_6$
	1	0	0	0	0	0

Answer

As we see that the given assignment is following both the conditions of axiomatic approach of probability
Condition (i): Each of the number p($\omega_i$) is 0 ≤p($\omega_i$)≤ 1
Condition (ii) Sum of the probabilities = 1 + 0 + 0 + 0 + 0 + 0 = 1
Therefore, the assignment is valid.

View full question & answer→

Question 581 Mark

Let a sample space be S = {$\omega_1, \omega_2,..., \omega_6$}.Which of the following assignments of probabilities to each outcome are valid?

Outcomes	$\omega_1$	$\omega_2$	$\omega_3$	$\omega_4$	$\omega_5$	$\omega_6$
	$\frac 16$	$\frac 16$	$\frac 16$	$\frac 16$	$\frac 16$	$\frac 16$

Answer

Given assignment is following both the conditions to be valid.
Condition (i): Each of the number p($\omega_i$) is positive and less than one.
Condition (ii): Sum of probabilities
$= \frac16 + \frac16 + \frac16+ \frac16+ \frac16+\frac16 = 1$
Therefore, the assignment is valid

View full question & answer→

Question 591 Mark

Consider the experiment of rolling a die. Let A be the event getting a prime number, B be the event getting an odd number. Write the sets representing the event not A.

Answer

Here the sample space S = {1, 2, 3, 4, 5, 6}, A = {2, 3, 5} and B = {1, 3, 5}
‘not A’ = A′ = {1,4,6}

View full question & answer→

Question 601 Mark

Consider the experiment of rolling a die. Let A be the event getting a prime number, B be the event getting an odd number. Write the sets representing the event A but not B.

Answer

Here sample space S = {1, 2, 3, 4, 5, 6}, A = {2, 3, 5} and B = {1, 3, 5}
‘A but not B’ = A – B = {2}

View full question & answer→

Question 611 Mark

Consider the experiment of rolling a die. Let A be the event getting a prime number, B be the event getting an odd number. Write the sets representing the event A and B.

Answer

Here sample space S = {1, 2, 3, 4, 5, 6}, A = {2, 3, 5} and B = {1, 3, 5}
‘A and B’ = A $\cap$ B = {3,5}

View full question & answer→

Question 621 Mark

Consider the experiment of rolling a die. Let A be the event getting a prime number, B be the event getting an odd number. Write the sets representing the event A or B

Answer

Here sample space s= {1, 2, 3, 4, 5, 6}, A = {2, 3, 5} and B = {1, 3, 5}
‘A or B’ = A $\cup$ B = {1, 2, 3, 5}

View full question & answer→

Question 631 Mark

Consider the experiment in which a coin is tossed repeatedly until a head comes up. Describe the sample space.

Answer

The experiment will continue until we get a head.
So, head may come up on the first toss and the outcome will be H,
If first toss is tail and 2nd toss is head outcome will be TH,
In the same way, if first and second toss is tails and third toss is a head, outcome will be TTH and so on till head is obtained.
Hence, the desired sample space is S= {H, TH, TTH, TTTH, TTTTH,...}

View full question & answer→

Question 641 Mark

A coin is tossed. If it shows head, we draw a ball from a bag consisting of $3$ blue and $4$ white balls; if it shows tail we throw a die. Describe the sample space of this experiment.

Answer

Let us denote three blue balls by $B_1, B_2, B_3 $nand four white balls by $W_1, W_2, W_3, W_4.$
Then a sample space of the experiment is
$S = \{ HB_1, HB_2, HB_3, HW_1, HW_2, HW_3, HW_4, T1, T2, T3, T4, T5, T6\}.$
Here $HB1$ means head on the coin and blue ball B1 is drawn, HW1 means head on the coin and white ball $W1$ is drawn. Similarly, $T1$ means tail on the coin and the number $1$ on the die.

View full question & answer→

Question 651 Mark

Specify appropriate sample space. A person is noting down the number of accidents along a busy highway during a year.

Answer

The number of accidents along a busy highway during the year of observation can be either 0 (for no accident) or 1 or 2, or some other positive integer.
Thus, a sample space of the event is given by
S = {0, 1, 2, ...}

View full question & answer→

Question 661 Mark

Specify appropriate sample space. A boy has a 1 rupee coin, a 2 rupee coin and a 5 rupee coin in his pocket. He takes out two coins out of his pocket, one after the other.

Answer

Let P denotes a 1 rupee coin, Q denotes a 2 rupee coin and R denotes a 5 rupee coin.
The first coin he takes out of his pocket may be any one of the three coins P, Q or R.
Corresponding to P, the second draw may be Q or R. So the result of two draws may be PQ or PR.
Similarly, corresponding to Q, the second draw may be R or P.
Therefore, the outcomes may be QP or QR.
Lastly, corresponding to R, the second draw may be P or Q.
So, the outcomes may be RP or RQ.
Thus, after combining all of the above outcomes, the sample space is S = {PQ, PR, QR, QP, RP, RQ}

View full question & answer→

Question 671 Mark

In a relay race there are five teams $A, B, C, D$ and $E.$ What is the probability that $A, B$ and $C$ are first three to finish $($in any order$) ($Assume that all finishing orders are equally likely$)$

Answer

If we consider the sample space consisting of all finishing orders in the first three places, we will have $^5P_3, = \frac {5!}{(5-3)!}$ = $5 \times 4 \times 3$ = 60 = total number of outcomes
$A, B$ and $C$ are the first three finishers. There will be 3! arrangements for $A, B$ and $C.$ Therefore, the sample points corresponding to this event will be $3! = 6 =$ total number of favourable outcomes.
So, $P(A, B$ and $C$ are first three to finish$) = \frac{Total ~number~ of ~outcomes}{Total ~number~ of favourable~ outcomes}$$\frac {3!}{60} = \frac {6}{60} = \frac {1}{10}$

View full question & answer→

Question 681 Mark

In a relay race there are five teams $A, B, C, D$ and $E.$ What is the probability that $A, B$ and $C$ finish first, second and third, respectively.

Answer

We have to find the the probability that $A, B$ and $C$ finish first, second and third, respectively.
If we consider the sample space consisting of all finishing orders in the first three places, we will have $^5P_3$,
i.e., $\frac {5!}{(5-3)!} = 5 \times 4 \times 3 = 60$ sample points, each with a probability of $\frac 1{60}$.
A, B and C finish first, second and third, respectively. There is only one finishing order for this, i.e., $ABC$.
Thus $P(A, B$ and $C$ finish first, second and third respectively$) = \frac 1{60}$.

View full question & answer→

Question 691 Mark

On her vacations, Veena visits four cities (A, B, C, and D) in random order. What is the probability of A just before B?

Answer

We have to find the value the probability of A just before B
The number of arrangements (orders) in which Veena can visit four cities A, B, C, or D is 4! i.e., 24.Therefore, n (S) = 24.
Since the number of elements in the sample space of the experiment is 24 all of these outcomes are considered to be equally likely. A sample space for the experiment is
S = {ABCD, ABDC, ACBD, ACDB, ADBC, ADCB,
BACD, BADC, BDAC, BDCA, BCAD, BCDA,
CABD, CADB, CBDA, CBAD, CDAB, CDBA,
DABC, DACB, DBCA, DBAC, DCAB, DCBA}
Let G be the event ‘Veena visits A either first or second’.
Here G = {ABCD, ABDC, CABD, CDAB, DABC, DCAB}
Therefore, P(G) = $\frac {n(G)}{n(S)} = \frac {6}{24} = \frac 1{4}$

View full question & answer→

Question 701 Mark

On her vacations Veena visits four cities (A, B, C and D) in a random order. What is the probability of A either first or second?

Answer

The number of arrangements (orders) in which Veena can visit four cities A, B, C, or D is 4! i.e. 24.
Therefore, n (S) = 24.
Sample space for the experiment is given by
S = {ABCD, ABDC, ACBD, ACDB, ADBC, ADCB
BACD, BADC, BDAC, BDCA, BCAD, BCDA
CABD, CADB, CBDA, CBAD, CDAB, CDBA
DABC, DACB, DBCA, DBAC, DCAB, DCBA}
Let G be the event ‘Veena visits A either first or second’.
Here G = {ABCD, ABDC, ADBC, ACDB, ACBD, ADCB
BACD, BADC, CABD, CADB, DABC, DACB}
n(G) = 12
Therefore, P(G) = $\frac {n(G)}{n(S)} = \frac {12}{24} = \frac 1{2}$

View full question & answer→

Question 711 Mark

On her vacations Veena visits four cities (A, B, C and D) in a random order. What is the probability of A first and B last?

Answer

The number of arrangements (orders) in which Veena can visit four cities A, B, C, or D is 4! i.e. 24.
Therefore, total number of elements n (S) = 24.
Sample space for the experiment is given by
S = {ABCD, ABDC, ACBD, ACDB, ADBC, ADCB
BACD, BADC, BDAC, BDCA, BCAD, BCDA
CABD, CADB, CBDA, CBAD, CDAB, CDBA
DABC, DACB, DBCA, DBAC, DCAB, DCBA}
Let G be the event ‘Veena visits A first and B Last’.
Here G = {ACDB, ADCB}
n(G) = 2
Therefore, P(G) = $\frac {n(G)}{n(S)} = \frac 2{24} = \frac 1{12}$

View full question & answer→

Question 721 Mark

On her vacations Veena visits four cities (A, B, C and D) in a random order. What is the probability of A before B and B before C?

Answer

The number of arrangements (orders) in which Veena can visit four cities A, B, C, or D is 4! i.e. 24.
Therefore, total number of outcomes n (S) = 24.
Sample space for the experiment is given by
S = {ABCD, ABDC, ACBD, ACDB, ADBC, ADCB
BACD, BADC, BDAC, BDCA, BCAD, BCDA
CABD, CADB, CBDA, CBAD, CDAB, CDBA
DABC, DACB, DBCA, DBAC, DCAB, DCBA}
Let the event ‘Veena visits A before B and B before C’ be denoted by F.
Here F = {ABCD, DABC, ABDC, ADBC}
n(F) = 4
Therefore, P(F) = $\frac {n(F)}{n(S)} = \frac 4{24} = \frac 16$

View full question & answer→

Question 731 Mark

On her vacations Veena visits four cities (A, B, C and D) in a random order. What is the probability of A before B?

Answer

The number of arrangements (orders) in which Veena can visit four cities A, B, C, or D is 4! i.e. 24.

Therefore, total number of outcomes, n (S) = 24.
Sample space for the experiment is given by
S = {ABCD, ABDC, ACBD, ACDB, ADBC, ADCB
BACD, BADC, BDAC, BDCA, BCAD, BCDA
CABD, CADB, CBDA, CBAD, CDAB, CDBA
DABC, DACB, DBCA, DBAC, DCAB, DCBA}
Let the event ‘she visits A before B’ be denoted by E
Therefore, E = {ABCD, CABD, DABC, ABDC, CADB, DACB
ACBD, ACDB, ADBC, CDAB, DCAB, ADCB}
n(E) =12
Thus P(E) = $\frac {n(E)}{n(S)} =\frac {12}{24} = \frac 12$

View full question & answer→

Question 741 Mark

A committee of two persons is selected from two men and two women. What is the probability that the committee will have two men?

Answer

It is given that, there are two men and two women.
Now, a committee of two persons is selected.
$\therefore$ $n (S) =\ ^4C_2 = \frac{4 \times 3}{2} = 6$
Let E be the event that two men in the committee
$\therefore n(E) =\ {^2C_2} = 1$
Therefore, $ P (E) =$ $\frac{1}{6}$

View full question & answer→

Question 751 Mark

A committee of two persons is selected from two men and two women. What is the probability that the committee will have one man?

Answer

It is given that, there are two men and two women.
Now, a committee of two persons is selected.
$\therefore$ $n (S) =\ ^4C_2 =$ $\frac{4 \times 3}{2}$
Let E be the event that one man is in the committee
$\therefore$ $n (E) =\ ^2C_1 \times\ ^2C_1$
= 2 $\times $ 2 = 4
Therefore, P(E) = $\frac{4}{6}=\frac{2}{3}$

View full question & answer→

Question 761 Mark

A committee of two persons is selected from two men and two women. What is the probability that the committee will have no man?

Answer

We have to find the probability that the committee will have no man
Given, there are two men and two women.
Now, a committee of two persons is selected.
$\therefore$ $n (S) =\ ^4C_2 =$ $\frac{4 \times 3}{2}$ = 6
Let E be the event that no man is to be in the committee
$\therefore$ $n (E) =\ ^2C_2 = 1$ [Only women will be in the committee]
$\therefore$ $P (E) = $$\frac{1}{6}$

View full question & answer→

Question 771 Mark

Two students Anil and Ashima appeared in an examination. The probability that Anil will qualify the examination is 0.05 and that Ashima will qualify the examination is 0.10. The probability that both will qualify the examination is 0.02. Find the probability that only one of them will qualify the examination.

Answer

Let E and F be the events that Anil and Ashima will qualify the examination, respectively.
Given that
P(E) = 0.05, P(F) = 0.10 and P(E $\cap$ F) = 0.02.
The event only one of them will qualify the examination is same as the event either (Anil will qualify, and Ashima will not qualify) or (Anil will not qualify and Ashima will qualify) i.e., E $\cap$ F´ or E´ $\cap$ F, where E $\cap$ F´ and E´ $\cap$ F are mutually exclusive
Therefore,
P(only one of them will qualify)
= P(E $\cap$ F´ or E´ $\cap$ F)
= P(E $\cap$ F´) + P(E´ $\cap$ F) - P [(E $\cap$ F´) $\cap$ P(E´ $\cap$ F)] [by general addition rule and also P [(E $\cap$ F´) $\cap$ P(E´ $\cap$ F)]= 0]
= P (E) – P(E $\cap$ F) + P(F) – P (E $\cap$ F)
= 0.05 – 0.02 + 0.10 – 0.02 = 0.11

View full question & answer→

Question 781 Mark

Two students Anil and Ashima appeared in an examination. The probability that Anil will qualify the examination is 0.05 and that Ashima will qualify the examination is 0.10. The probability that both will qualify the examination is 0.02. Find the probability that at least one of them will not qualify the examination.

Answer

Let E and F denote the events that Anil and Ashima will qualify the examination, respectively.
Given that
P(E) = 0.05, P(F) = 0.10 and P(E $\cap$ F) = 0.02.
P(at least one of them will not qualify)
= 1 - P(both of them will qualify)
= 1 - P(E $\cap$ F)
= 1 - 0.02
= 0.98

View full question & answer→

Question 791 Mark

Two students Anil and Ashima appeared in an examination. The probability that Anil will qualify the examination is 0.05 and that Ashima will qualify the examination is 0.10. The probability that both will qualify the examination is 0.02. Find the probability that 'Both Anil and Ashima will not qualify the examination.

Answer

Let E and F denote the events that Anil and Ashima will qualify the examination, respectively.
Given that
P(E) = 0.05, P(F) = 0.10 and P(E $\cap$ F) = 0.02.
Now,the event ‘both Anil and Ashima will not qualify the examination’ may be expressed as E´ $\cap$ F´.
Since, E´ is ‘not E’, i.e., Anil will not qualify the examination and F´ is ‘not F’, i.e., Ashima will not qualify the examination.
Also E´ $\cap$ F´ = (E $\cup$ F)´ (by Demorgan's Law)
So, P(E´ $\cap$ F´) = P(E $\cup$ F)´
Applying the formula,
P(E $\cup$ F) = P(E) + P(F) – P(E $\cup$ F)
or P(E ∪ F) = 0.05 + 0.10 – 0.02 = 0.13
Therefore P(E´ $\cap$ F´) = P(E $\cup$ F)´ = 1 – P(E $\cup$ F) = 1 – 0.13 = 0.87

View full question & answer→

Question 801 Mark

One card is drawn from a well shuffled deck of 52 cards. If each outcome is equally likely, calculate the probability that the card will be not a black card.

Answer

When a card is drawn from a well shuffled deck of 52 cards, the number of possible outcomes = 52.
Let C be the event ‘card drawn is black card’
As we know that total number of black cards = 26
So P(C) = $\frac {26}{52} = \frac 12$
Thus, probability of a black card = $\frac 12$
The event ‘card drawn is not a black card’ may be denoted as C′ or ‘not C’.
We know that P(not C) = 1 – P(C) = $1 - \frac 12 = \frac 12$
Therefore, probability of not a black card = $\frac 12$

View full question & answer→

Question 811 Mark

One card is drawn from a well shuffled deck of 52 cards. If each outcome is equally likely, calculate the probability that the card will be not a diamond.

Answer

When a card is drawn from a well shuffled deck of 52 cards, the number of possible outcomes = 52.
Let A be the event 'the card drawn is a diamond'
We know that total number of diamond cards = 13.
Therefore, P(A)$= \frac {13}{52} = \frac 14$
So the event ‘card drawn is not a diamond’ may be denoted as A' or ‘not A’
Now P(not A) = 1 – P(A) $= 1-\frac14 = \frac 34 $

View full question & answer→

Question 821 Mark

One card is drawn from a well shuffled deck of 52 cards. If each outcome is equally likely, calculate the probability that the card will be a black card (i.e., a club or, a spade).

Answer

When a card is drawn from a well shuffled deck of 52 cards, the number of possible outcomes is 52.
Let C be the event ‘card drawn is black card’.
Since total number of black cards = 26
So, P(C) = $\frac {26}{52} = \frac 12$
Thus, probability of a black card = $\frac 12$

View full question & answer→

Question 831 Mark

One card is drawn from a well shuffled deck of 52 cards. If each outcome is equally likely, calculate the probability that the card will be not an ace.

Answer

When a card is drawn from a well shuffled deck of 52 cards, the number of possible outcomes is 52.
Let A be the event ‘Card drawn is an ace’.
Since there are 4 cards of ace, so P(B) = $\frac{4}{52}$
We know that P($\bar B$) = 1 – P(B) = $= 1 - \frac 4{52} = 1 - \frac 1{13} = \frac {12}{13}$

View full question & answer→

Question 841 Mark

One card is drawn from a well shuffled deck of 52 cards. If each outcome is equally likely, calculate the probability that the card will be 'a diamond'

Answer

When a card is drawn from a well shuffled deck of 52 cards,
Total number of possible outcomes = 52.
Let A be the event 'the card drawn is a diamond'
Clearly the number of elements in set A is 13.
Total number of favorable outcomes = 13
Therefore, P(A) $= \frac {13}{52} = \frac 14$
i.e. probability of being a diamond card = $\frac 14$

View full question & answer→

Question 851 Mark

Two coins (a one rupee coin and a two rupee coin) are tossed once. Find a sample space.

Answer

Since the coins are distinguishable so we can speak of the first coin and the second coin. Since either coin can turn up Head (H) or Tail (T), the possible outcomes may be
Heads on both the coins = (H,H) = HH
Head on first coin and Tail on the other = (H,T) = HT
Tail on first coin and Head on the other = (T,H) = TH
Tail on both coins = (T,T) = TT
Combination of all the above outcomes will give us the sample space
Thus, the sample space is S = {HH, HT, TH, TT}

View full question & answer→

Generate Paper Free All Probability topics

1 Marks Question - Page 2 - Maths STD 11 Science Questions - Vidyadip