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Multiple choice maths introduction of probability theory types of events some more terms in probability events and its algebra

Tickets numbered from $1$ to $30$ are mixed up and then a ticket is drawn at random. What is the probability that the drawn ticket has a number which is divisible by both $2$ and $6$?

  1. $\dfrac{1}{2}$
  2. $\dfrac{2}{5}$
  3. $\dfrac{8}{15}$
  4. $\dfrac{1}{6}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Here, $S={1,2,3,4,5,6,.....29,30}$


Let $E$ be the event of number divisible by both $2$ and $6$.

$E={6,12,18,24,30}$

$P(E)=\dfrac{n(E)}{n(S)}=\dfrac{5}{30}=\dfrac{1}{6}$

$\therefore$   the probability that the drawn ticket has a number which is divisible by both $2$ and $6$ is $\dfrac{1}{6}$

Multiple choice maths introduction of probability theory types of events some more terms in probability events and its algebra

The number of ways in which $6$ men can be arranged in a row, so that three particular men are consecutive, is 

  1. $4! \times 3!$
  2. $4!$
  3. $3! \times 3!$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
To arrange $6$ men in a row such three particular men are consecutive $M _1M _2M _3M _4M _5M _6$
three men to be consecutive lets make $3$ men in a Group
$(M _1M _2M _3)$ $M _4M _5M _6$
Total no. of ways of arranging them is $4!\times3!$.
Multiple choice maths introduction of probability theory types of events some more terms in probability events and its algebra

If A and B are such events that $P(A)>0$ and $ P(B)\neq 1$ then $P\left(\dfrac{\bar{A}}{\bar{B}}\right)$ is equal to-

  1. $1-P\left(\dfrac{A}{B}\right)$
  2. $1-P\left(\dfrac{\bar{A}}{B}\right)$
  3. $\dfrac{1-P(A\cup B)}{P(\bar{B})}$
  4. $None$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using the definition of conditional probability, P(A_bar/B_bar) = P(A_bar intersection B_bar) / P(B_bar). By De Morgan's law, A_bar intersection B_bar = (A union B)_bar, so the numerator is 1 - P(A union B). Thus, the expression is (1 - P(A union B)) / P(B_bar).

Multiple choice maths introduction of probability theory types of events some more terms in probability events and its algebra

4 normal distinguishable dice are rolled once. The number of possible outcomes in which at least one dice shows up 2?

  1. 216

  2. 648

  3. 625

  4. 671
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
The number of possible outcomes in which atleast one dice shows up $2$ is:
$\begin{array}{l} =(6\times 6\times 6\times 6)-(5\times 5\times \times 5\times 5) \\= 1296-625=671 \end{array}$

Hence, the correct option is $D$
Multiple choice maths introduction of probability theory types of events some more terms in probability events and its algebra

$8$ players compete in a tournament, every one plays everyone else just once. The winner of a game gets $1$, the loser $0$ or each gets $\dfrac{1}{2}$ if the game is drawn. The final result is that every one gets a different score and the player playing placing second gets the same as the total of four bottom players.The total score of all the players is

  1. $28$
  2. $21$
  3. $20$
  4. $22$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The players get $1$ for winning 

and $\dfrac 12$ for a draw 
So the sum of points of the players for one match is $1$
Hence the total score for $^8C _2=\dfrac{8\times7}2$ matches is $28$
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Multiple choice maths introduction of probability theory types of events some more terms in probability events and its algebra

A fair die is thrown 3 times . The chance that sum of three numbers appearing on the die is less than 11 , is equal to -

  1. $\dfrac{1}{2}$
  2. $\dfrac{2}{3}$
  3. $\dfrac{1}{6}$
  4. $\dfrac{5}{8}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
From given, we have,

Sum 3:$ (1, 1, 1)$ ==> Contributing only $1$ distinct triplet.

Sum 4: $(1, 1, 2)$ ==> Contributing $3$ distinct triplets

Sum 5: $(1, 2, 2) $and $(1, 1, 3)$ ==> Contributing $6$ distinct triplets

Sum 6: $(1, 1, 4), (1, 2, 3)$ and $(2, 2, 2)$ => Together contributing $10 $distinct triplets

Sum 7: $(1, 1, 5), (1, 2, 4), (1, 3, 3) $and $(2, 2, 3)$ => Together contributing $15$ distinct triplets.

Sum 8: $(1, 1, 6), (1, 2, 5), (1, 3, 4), (2, 3, 3)$ and$ (2, 4, 2) $==> Together contributing $21$ distinct triplets.

Sum 9: $(1, 2, 6), (1, 3, 5), (1, 4, 4), (2, 3, 4), (2, 5, 2)$ and $(3, 3, 3)$ ==> Together contributing $25$ distinct triplets.

Sum 10: $(1, 3, 6), (1, 4, 5), (2, 2, 6), (2, 3, 5), (2, 4, 4)$ and $(3, 3, 4) $==> Together contributing $27$ distinct triplets.

Therefore number of favorable cases $= 1+ 3 + 6 + 10 + 15 + 21 + 25 + 27 = 108.$

Therefore, probability $= \dfrac{108}{216} =\dfrac{1}{2}$
Multiple choice maths introduction of probability theory types of events some more terms in probability events and its algebra

The probability that a number selected at random from the numbers $1,2,3.......15$ is a multiple of $4$ is 

  1. $\dfrac{4}{15}$
  2. $\dfrac{2}{15}$
  3. $\dfrac{1}{15}$
  4. $\dfrac{1}{5}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

From Number$ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15$

From 1 to 15, Multiples of 4 are 4, 8, 12 only

So Probability$ = \dfrac {Count \ of \ No. \ which \ are \ multiple \ of \ 4}{Total \ Number \ given}$

$ Prob. = \dfrac{3}{15} = \dfrac{1}{5}$

Option D is correct

Multiple choice maths introduction of probability theory types of events some more terms in probability events and its algebra

Three letters, to each of which corresponds an envelope, are placed in the envelopes at random. The probability that all the letters are not placed in the right envelopes, is

  1. $\dfrac{1}{6}$
  2. $\dfrac{5}{6}$
  3. $\dfrac{1}{3}$
  4. $\dfrac{2}{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Three letters can be placed in 3 envelopes in $3!$ ways, whereas there is only one way of placing them in their right envelopes.
So, probability that all the letters are placed in the right envelopes$=\dfrac{1}{3!}$
Hence, required probability$=1-\dfrac{1}{3!}=\dfrac{5}{6}$

Multiple choice maths introduction of probability theory types of events some more terms in probability events and its algebra

A coin is tossed and a single $6$-sided die is rolled. Find the probability of landing on the tail side of the coin and rolling $4$ on the die.

  1. $\dfrac{1}{12}$
  2. $\dfrac{6}{5}$
  3. $\dfrac{4}{3}$
  4. $\dfrac{3}{4}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$P$ (tail) $=$ $\dfrac{1}{2}$ and $P(4) =$ $\dfrac{1}{6}$

$P$ (tail and $4$) $=$ $P$(tail) $. P(4)$
$=$$\cfrac{1}{2}\times \cfrac{1}{6}$ $=$ $\cfrac{1}{12}$

Multiple choice maths introduction of probability theory types of events some more terms in probability events and its algebra

The probability of getting number less than or equal to $6$, when a die is thrown once, is

  1. An impossible event

  2. A sure event

  3. An exhaustive event

  4. A complementary event
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The probability of getting number less than $6$, when a die is thrown once, is a sure event.
Because, once a die is thrown, sample space $= {1, 2, 3, 4, 5, 6}$
There is a possible event for getting number less than $6$ as outcomes can be $1, 2, 3, 4, 5$.