Tag: some more terms in probability

Questions Related to some more terms in probability

If $P(A) = 0$, then the event $A$

  1. Will never happen

  2. Will always happen

  3. May happen

  4. May not happen


Correct Option: A
Explanation:

The probability is the possibility of an event happening when the probability of an event is zero, it means the event will never occur.

The probability of a sure event (or certain event) is ____

  1. $0$

  2. $1$

  3. $2$

  4. $3$


Correct Option: B
Explanation:

Probability of event in $E$, $ P\left( E \right) =\dfrac { \text {number of events in E} }{ \text {Tota number of possible events} } $

If the number of desirable events is equal to the total number of possible events, then probability of an event which is certain or sure to happen $ = 1$
Such an event is called certaind event or sure event.

The probability of an event that is certain to happen is ____?

  1. $1$

  2. $2$

  3. $3$

  4. $4$


Correct Option: A
Explanation:

The event that is sure to happen is called a certain event and probability of such an event is $1$ as this event is bound to happen.

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. $\dfrac{1}{6}$


Correct Option: D
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 $\dfrac{1}{6}$

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


Correct Option: A
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 $4!\times3!$.

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. $None$


Correct Option: C

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


Correct Option: D
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 $D$

$8$ 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. $22$


Correct Option: A
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 $28$
'

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. $\dfrac{5}{8}$


Correct Option: A
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 $= \dfrac{108}{216} =\dfrac{1}{2}$

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. $\dfrac{1}{5}$


Correct Option: D
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