3 Marks Question

Question 13 Marks

Two different dice are thrown together. Find the probability that the numbers obtained
$(i)$ have a sum less than $7$
$(ii)$ have a product less than $16$
$(iii)$ is a doublet of odd numbers.

Answer

We know that the total number of outcome when two dice are thrown to gether is $36$
$(i)$ Have a sum less than $7$
The favourable outcomes are
$=\{(1,1),(1,2),(1,3),(1,4),(1,5),(2,1),$
$(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),$
$(4,1),(4,2),(5,1)\}$
Number of favourable outcomes $=15$
So, probability of getting a sum less than $7$
$=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{15}{36}$
$=\frac{5}{12}$
$(ii)$ Have a product less than $16 $
The favourable outcomes are
$=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),$
$(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),$
$(3,3),(3,4),(3,5),(4,1),(4,2),(4,3),(5,1),$
$(5,2),(5,3),(6,1),(6,2)\}$
Number of favourable outcomes $=25$
So, probability will be
$=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{25}{36}$
$(iii) A$ doublet of odd numbers.
The favourable outcomes are
$=\{(1,1),(3,3),(5,5)\}$
Number of favourable outcomes $=3$
So, probability will be
$=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{3}{36}$
$=\frac{1}{12}$

View full question & answer→

Question 23 Marks

A bag contains $15$ white and some black balls. If the probability of drawing a black ball from the bag is thrice that of drawing a white ball, find the number of black balls in the bag.

Answer

Let us assume that the number of black ball $=x$ Number of white balls $= 15$
$P(\text { Black Ball })=3 \times P(\text { White Balls })$
$\Rightarrow \frac{x}{15+x}=3 \times \frac{15}{15+x}$
$\Rightarrow x=3 \times 15$
$\Rightarrow x=45$
Therefore, number of black balls are $45 .$

View full question & answer→

Question 33 Marks

In a single throw of a pair of different dice, what is the probability of getting $(i)$ a prime number on each dice? $(ii)$ a total of $9$ or $11$ ?

Answer

We know that the total number of outcomes on throwing a pair of dice $=6 \times 6=36$
$(i)$ Let $A$ be the event of getting a prime number on each dice.
So, the favourable outcomes
$=\left\{(2,2),(2,3),(2,5),(3,2),(3,3),(3,5),(5,2),(5,3),(5,5) \right\}$
Number of favourable outcomes $=9$
Now,
$P(A)=\frac{\text { Number of favourable outcomes }}{\text { Total possible outcomes }}=\frac{9}{36}=\frac{1}{4}$
Thus, the probability of getting a prime number on each dice is $\frac{1}{4}$.
$(ii)$ Let $B$ be the event of getting a total of $9$ or $11$ .
So, the favourable outcomes
$=\{(3,6),(4,5),(5,4),(6,3),(5,6),(6,5)\}$
Number of favourable outcomes $=9$
Now, $P(B)=\frac{6}{36}=\frac{1}{6}$
Thus, the probability of getting a total of $9$ or $11$ is $\frac{1}{6}$.

View full question & answer→

Question 43 Marks

Three different coins are tossed together. Find the probability of getting
(i) Exactly two heads
(ii) At least two heads
(iii) at least two tails

Answer

The possible outcomes, when three coins are tossed together are
{HHH, HHT, HTT, HTH, THH, TTH, THT, TTT}
Therefore, total number of possible outcomes $=8$
(i) Favorable outcomes of exactly two heads are HHT, HTH, THH
Therefore, total number of possible outcomes $=3$
Probability of getting exactly two head
$
=\frac{\text { Total number of favourable outcomes }}{\text { Total possible outcomes }}=\frac{3}{8}
$
(ii) Favorable outcomes of at least two heads are HHH, HHT, HTH, THH
Therefore, total number of possible outcomes $=4$
Probability of getting at least two heads
$
=\frac{\text { Total number of favourable outcomes }}{\text { Total possible outcomes }}=\frac{4}{8}=\frac{1}{2}
$
(iii) Favorable outcomes of at least two tails are HHT, TTH, THT, TTT
Therefore, total number of possible outcomes $=4$
Probability of getting at least two tails
$
=\frac{\text { Total number of favourable outcomes }}{\text { Total possible outcomes }}=\frac{4}{8}=\frac{1}{2}
$

View full question & answer→

Question 53 Marks

All kings, queens and aces are removed from a pack of 52 cards. The remaining cards are well shuffled and then a card is drawn from it. Find the probability that the drawn card is
(i) A black face card
(ii) A red card

Answer

In a pack of 52 playing cards, there are 4 kings ( 2 red +2 black), 4 queens ( 2 red +2 black) and aces ( 2 black +2 red).
So, when all the kings, queens and aces are removed from a pack of 52 cards, the remaining cards are $52-12=40$
Now, in a standard deck of 52 playing cards, there are four suits: clubs, diamonds, hearts and spades. Each suit has one Jack, Queen, and King as the face cards. Hence there are 12 face cards in a deck of 52 playing cards.
So, after removing all the kings and queens cards from the pack, there will be $4(2$ red +2 black) face cards remaining in the deck of 40 cards.
(i) Probability of getting a black face card is $=\frac{2}{40}=\frac{1}{20}$.
(ii) In a deck of 52 cards, 26 cards are black cards and 26 cards are red cards. When all the kings, queens and aces are removed, out of the m 2 kings, 2 queens and 2 aces are red cards. So, the total number of red cards remaining in the deck of cards is $26-6=20$.
Probability of getting a red card
$
=\frac{20}{40}=\frac{1}{2}
$

View full question & answer→

Question 63 Marks

A game consists of tossing a coin three times and noting its outcome each time. Hanif wins if he gets three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.

Answer

When a coin is tossed three times, then the possible outcomes are:
{(THH), (TTT), (TTH), (THT), (HHH), (HTT), (HTH), (HHT)}
$\therefore$ Total outcomes $=8$
Favorable outcome for three heads or three tails are (HHH), (TTT) i.e. 2 outcomes.
Probability of Hanif winning the game
$=\frac{\text { Favourable outcome for three heads or three tails }}{\text { Total outcome }}$

$
=\frac{2}{8}=\frac{1}{4}
$
Clearly, probability of Hanif losing the game
$=1-$ Probability of winning
$
\Rightarrow 1-\frac{1}{4} \Rightarrow \frac{3}{4}
$
Therefore, the probability that Hanif will lose the game is $\frac{3}{4}$.

View full question & answer→

Question 73 Marks

Two dice are rolled once. Find the probability of getting such numbers on two dice, whose product is a perfect square.

Answer

When two dice are rolled, then the possible outcomes are:
$
\left(\begin{array}{llllll}
(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)
\end{array}\right)=36 \text { outcomes }
$
Favorable outcomes for the numbers on two dice, whose product is a perfect square are $\{(1,1)(1,4)(2,2)(3,3)(4,1)(4,4)(5,5)(6,6)\}$, i.e. 8 outcomes.
Therefore, probability of getting such numbers on two dice, whose product is a perfect square
$
=\frac{\text { Favourable outcome }}{\text { Total outcome }}=\frac{8}{36}=\frac{2}{9}
$

View full question & answer→

Question 83 Marks

A die is rolled once. Find the probability of getting :
(i) an even prime number.
(ii) a number greater than 4.
(iii) an odd number.

Answer

(i) Out of 1 to 6 numbers, the only even prime number is 2 .
The profitability of getting an even prime number $=\frac{1}{6}$
(ii) Out of 1 to 6 numbers, numbers greater than 4 are 2
The probability of getting number greater than
$
4=\frac{2}{6}=\frac{1}{3}
$
(iii)Out of 1 to 6 numbers, numbers which are odd $=1,3,5$
The probability of getting an odd number $=\frac{3}{6}=\frac{1}{2}$

View full question & answer→

Maths 3 Marks Question

Take a timed test