Exercise 8.3 - Chapter 8 Statistics 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

$\mathbf{E x} 8.3$
Question 1.
Write the sample space for tossing three coins using tree diagram.

Question 2.
Write the sample space for selecting two balls from a bag containing 6 balls numbered 1 to 6 (using tree diagram).
Solution:

$\mathrm{S}=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2)$ $(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5)$, $(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}$
Question 3.
If $\mathrm{A}$ is an event of a random experiment such that $\mathrm{P}(\mathrm{A}): \mathrm{P}(\overline{\mathbf{A}})=17: 15$ and $\mathrm{n}(\mathrm{S})=640$ then find
(i) $\mathrm{P}(\overline{\mathbf{A}})$
(ii) $n(\mathrm{~A})$.
Solution:

$\mathrm{P}(\mathrm{A}): \mathrm{P}(\overline{\mathbf{A}})=17: 15$
(i) $\mathrm{P}(\mathrm{A})=\frac{17}{32}, \mathrm{P}(\overline{\mathrm{A}})=\frac{15}{32}$
(ii) $\mathrm{P}(\mathrm{A})=\frac{17}{32}=\frac{17 \times 20}{32 \times 20}=\frac{340}{640}$
$
\Rightarrow \quad n(\mathrm{~A})=340
$
Question 4.
A coin is tossed thrice. What is the probability of getting two consecutive tails?
Solution:
Outcomes $\{\mathrm{O}\}:\{(\mathrm{HHH}),(\mathrm{THH}),(\mathrm{HTH})$, (HHT), (HTT), (THT), (TTH), (TTT) $\}$
Two consecutive tails $\{\mathrm{F}\}:\{(\mathrm{HTT})$, (TTH), (TTT) $\}$
$
\begin{aligned}
& \mathrm{n}\{\mathrm{F}\}=3 \\
& \mathrm{n}\{\mathrm{O}\}=8 \\
& \Rightarrow \mathrm{P}=\frac{n\{\mathrm{~F}\}}{n\{\mathrm{O}\}}=\frac{3}{8}
\end{aligned}
$
Question 5.
At a fete, cards bearing numbers 1 to 1000 , one number on one card are put in a box. Each player selects one card at random and that card is not replaced. If the selected card has a perfect square number greater than 500 , the player wins a prize. What is the probability that
(i) the first player wins a prize
(ii) the second player wins a prize if the first has won?
Answer:
Sample space $=\{1,2,3, \ldots, 1000\}$
$\mathrm{n}(\mathrm{S})=1000$
(i) Let A be the event of setting square number greater than 500
$\mathrm{A}=\{529,576,625,676,729,784,841,900,961\}$
$\mathrm{n}(\mathrm{A})=9$
$\mathrm{P}(\mathrm{A})=\frac{n(\mathrm{~A})}{n(\mathrm{~S})}=\frac{9}{1000}$
The probability that the first player wins prize $=\frac{9}{1000}$
(ii) If the first player wins, the number is excluded for the second player.
$\mathrm{n}(\mathrm{A})=8$ and $\mathrm{n}(\mathrm{S})=999$
$\mathrm{P}(\mathrm{A})=\frac{n(A)}{n(S)}=\frac{8}{999}$
Probability the second player wins a prize $=\frac{8}{999}$

Question 6.
A bag contains 12 blue balls and $\mathrm{x}$ red balls. If one ball is drawn at random
(i) what is the probability that it will be a red ball?
(ii) If 8 more red balls are put in the bag, and if the probability of drawing a red ball will be twice that of the probability in (i), then find $\mathrm{x}$.
Solution:
$12 \rightarrow$ blue balls
$\mathrm{x} \rightarrow$ red balls
(i) $P($ red ball $)=\frac{x}{x+12}$
(ii) 8 red balls are added to the bag.
$\therefore 12 \rightarrow$ blue balls
$\mathrm{x}+8 \rightarrow$ red balls
$\therefore \mathrm{P}(\text { red ball })_{\text {new }}=\frac{x+8}{x+8+12}=\frac{x+8}{x+20}$
Given that $\mathrm{P}($ ii $)=2 \times \mathrm{P}(\mathrm{i})$
$\Rightarrow \frac{x+8}{x+20}=2 \times \frac{x}{x+12}$
$\Rightarrow(\mathrm{x}+8)(\mathrm{x}+12)=2 \mathrm{x}(\mathrm{x}+20)$
$\Rightarrow\left(\mathrm{x}^2+20 \mathrm{x}+96\right)=2 \mathrm{x}^2+40 \mathrm{x}$
$\Rightarrow \mathrm{x}^2+20 \mathrm{x}-96=0$
$\Rightarrow \mathrm{x}^2+24 \mathrm{x}-4 \mathrm{x}-96=0$
$\Rightarrow \mathrm{x}(\mathrm{x}+24)-4(\mathrm{x}+24)=0$
$\Rightarrow(\mathrm{x}-4)(\mathrm{x}+24)=0$
$\therefore \mathrm{x}=4$ (or) $\mathrm{x}=-24$
$x$ cannot be negative $\Rightarrow x=4$
Substituting $x=4$ in (i)
we get $P($ red ball $)=\frac{4}{4+12}=\frac{1}{4}$
Question 7.
Two unbiased dice are rolled once. Find the probability of getting
(i) a doublet (equal numbers on both dice)
(ii) the product as a prime number
(iii) the sum as a prime number
(iv) the sum as 1
Solution:
Doublet $=\{(1,1)(2,2)(3,3)(4,4)(5,5)(6,6)\}$
Total number of outcomes $=6 \times 6$

$
\mathrm{n}(\mathrm{S})=36
$
Number of favourable outcomes $=6$
$
\Rightarrow P(\text { doublet })=\frac{6}{36}=\frac{1}{6}
$
(ii) Number of favourable outcomes $=6$
as favourable outcomes $=(1,2),(2,1),(1,3),(3,1),(1,5)$, and $(5,1)$
$
\Rightarrow P(\text { prime number as product })=\frac{\not 6}{36_6}=\frac{1}{6}
$
(iii) Sum as prime numbers $=\{(1,1),(1,2),(2,3),(1,4),(1,6),(4,3),(5,6)\}$
Number of favourable outcomes $=7$
$\Rightarrow$ Probability $=\frac{7}{36}$
(iv) With two dice, minimum sum possible $=2$
$\therefore$ Prob $($ sum as 1$)=0[$ Impossible event $]$
Question 8.
Three fair coins are tossed together. Find the probability of getting
(i) all heads
(ii) atleast one tail
(iii) atmost one head
(iv) atmost two tails
Answer:
Three fair coins are tossed together
Sample spade $=\{\mathrm{HHH}, \mathrm{HHT}, \mathrm{HTH}, \mathrm{HTT}, \mathrm{THH}, \mathrm{THT}, \mathrm{TTH}, \mathrm{TTT}\}$ $\mathrm{n}(\mathrm{S})=8$
(i) Let A be the event of getting all heads
$
\mathrm{A}=\{\mathrm{HHH}\}
$
$
\mathrm{n}(\mathrm{A})=1
$
$
P(A)=\frac{n(A)}{n(S)}=\frac{1}{8}
$
(ii) Let $\mathrm{B}$ be the event of getting atleast one tail.
$\mathrm{B}=\{\mathrm{HHT}, \mathrm{HTH}, \mathrm{HTT}, \mathrm{THH}, \mathrm{THT}, \mathrm{TTH}, \mathrm{TTT}\}$
$\mathrm{n}(\mathrm{B})=7$
$
P(B)=\frac{n(B)}{n(S)}=\frac{7}{8}
$
(iii) Let $\mathrm{C}$ be the event of getting atmost one head
$\mathrm{C}=\{\mathrm{HTT}, \mathrm{THT}, \mathrm{TTH}, \mathrm{TTT}\}$
$
\mathrm{n}(\mathrm{C})=4
$

$
P(C)=\frac{n(C)}{n(S)}=\frac{4}{8}=\frac{1}{2}
$
(iv) Let D be the event of getting atmost two tails.
$\mathrm{D}=\{\mathrm{HTT}, \mathrm{TTT}, \mathrm{TTH}, \mathrm{THT}, \mathrm{THH}, \mathrm{HHT}, \mathrm{HTH}\}$
$
\begin{aligned}
& \mathrm{n}(\mathrm{D})=7 \\
& P(D)=\frac{n(D)}{n(S)}=\frac{7}{8}
\end{aligned}
$
Question 9.
Two dice are numbered 1,2,3,4,5,6 and 1,1,2,2,3,3 respectively. They are rolled and the sum of the numbers on them is noted. Find the probability of getting each sum from 2 to 9 separately.
Solution:
Dice 1
$
\mathrm{S}=\{1,2,3,4,5,6\}
$
Dice 2
$
\mathrm{S}=\{1,1,2,2,3,3\}
$
Total possible outcomes when they are rolled

$
\mathrm{n}(\mathrm{S})=36
$
Event of $\operatorname{sum}(2)=\mathrm{A}=\{(1,1),(1,1)\}$,
$
\mathrm{n}(\mathrm{A})=2, \mathrm{P}(\mathrm{A})=\frac{2}{36}
$
Event of sum 3 is B $=\{(1,2),(1,2),(2,1),(2,1)\}$
$
n(\mathrm{~B})=4, \mathrm{P}(\mathrm{B})=\frac{n(\mathrm{~B})}{n(\mathrm{~S})}=\frac{4}{36}
$
Event of sum 4 is $\mathrm{C}=\{(1,3),(1,3),(2,2),(2,2),(3,1)(3,1)\}$
$
\mathrm{n}(\mathrm{C})=6
$
$
P(C)=\frac{6}{36}
$
Event of getting the sum 5 is $\mathrm{D}=\{(2,3),(2,3),(3,2),(3,2),(4,1),(4,1)\}$ $\mathrm{n}(\mathrm{D})=6, \mathrm{P}(\mathrm{D})=\frac{6}{36}$

Event of getting the sum 6 is
$
\begin{aligned}
& \mathrm{E}=\{(3,3),(3,3),(4,2),(4,2),(5,1),(5,1)\} \\
& \mathrm{n}(\mathrm{E})=6, \mathrm{P}(\mathrm{E})=\frac{6}{36}
\end{aligned}
$
Event of getting the sum 7 is
$
\begin{aligned}
& \mathrm{F}=\{(4,3),(4,3),(5,2),(5,2),(6,1),(6,1)\} \\
& \mathrm{n}(\mathrm{F})=6 \\
& \mathrm{P}(\mathrm{F})=\frac{6}{36}
\end{aligned}
$
Event of getting the sum 8 is
$
\begin{aligned}
& \mathrm{G}=\{(5,3),(5,3),(6,2),(6,2)\} \\
& n(\mathrm{G})=4, \mathrm{P}(\mathrm{G})=\frac{4}{36}
\end{aligned}
$
Event of getting the sum 9 is
$
\begin{gathered}
\mathrm{H}=\{(6,3),(6,3), \mathrm{n}(\mathrm{H})=2 \\
\mathrm{P}(\mathrm{H})=\frac{2}{36} \\
\therefore \frac{2}{36}, \frac{4}{36}, \frac{6}{36}, \frac{6}{36}, \frac{6}{36}, \frac{6}{36}, \frac{4}{36}, \frac{2}{36}
\end{gathered}
$
Question 10.
A bag contains 5 red balls, 6 white balls, 7 green balls, 8 black balls. One ball is drawn at random from the bag. Find the probability that the ball is drawn
(i) white
(ii) black or red
(iii) not white
(iv) neither white nor black
Solution:

5 red 6 white 7 green 8 black total no. of balls $=5+6+7+8=26$
(i) $\quad$ Prob(white) $=\frac{6}{26_{13}}=\frac{3}{13}$
(ii) Prob(black or red) $=\frac{5+8}{26}=\frac{13}{26}=\frac{1}{2}$
(iii) Prob(not white) $=1-\operatorname{prob}$ (white)
$
\Rightarrow \quad=1-\frac{3}{13}=\frac{10}{13}
$
(iv) prob (neither her white or black) = 1-prob(white or black)
$
\begin{aligned}
& =1-\frac{6+8}{26}=1-\frac{14}{26} \\
& =\frac{26-14}{26}=\frac{2}{26}=\frac{6}{13}
\end{aligned}
$
Question 11.
In a box there are 20 non-defective and some defective bulbs. If the probability that a bulb selected at random from the box found to be defective is $3 / 8$ then, find the number of defective bulbs.
Solution:
Let number of defective bulbs be ' $x$ '
Total number of bulbs $=\mathrm{x}+20$
$
\begin{aligned}
& \text { Prob}(\text { defective })=\frac{x}{x+20}=\frac{3}{8} \Rightarrow 8 x=3(x+20) \\
& \Rightarrow 8 \mathrm{x}=3 \mathrm{x}+60 \\
& \Rightarrow 5 \mathrm{x}=60 \\
& \Rightarrow \mathrm{x}=12
\end{aligned}
$
$\therefore$ No.of defective bulbs are $=12$.
Question 12.
The king and queen of diamonds, queen and jack of hearts, jack and king of spades are removed from a deck of 52 playing cards and then well shuffled. Now one card is drawn at random from the remaining cards. Determine the probability that the card is
(i) a clavor
(ii) a queen of red card
(iii) a king of black card

Solution:

(i.e) remaining number of cards $=52-6=4613$
(i) $\mathrm{P}($ a clavor $)=\frac{13}{46}$
(ii) $\mathrm{P}($ queen of red card $)=0$ as both Queen of diamond and heart have been removed.
(iii) only $\mathrm{K}$ of clavor is in the deck
$\Rightarrow P($ king of black card $)=\frac{1}{46}$
Question 13.
Some boys are playing a game, in which the stone was thrown by them landing in a circular region (given in the figure) is considered as a win and landing other than the circular region is considered as a loss. What'is the probability to win the game?
4 feet

Solution:
Area of circular region $=\pi R^2=\pi(1)^2$
$=\pi$ sq.feet
Total area $=4 \times 3=12$ sq. feet
$
\begin{aligned}
\therefore \text { Prob(win the game) } & =\frac{\pi}{12}=\frac{3.14}{12} \\
& =\frac{314}{1200}=\frac{157}{600}
\end{aligned}
$
Question 14.
Two customers Priya and Amuthan are visiting a particular shop in the same week (Monday to Saturday). Each is equally likely to visit the shop on any one day as on another day. What is the probability that both will visit the shop on
(i) the same day
(ii) different days
(iii) consecutive days?

Solution:
Prob of Priya and Amurthan to visit shop on any
$
\text { day }=\frac{1}{6}
$
(i) $\operatorname{prob}\left(\right.$ visit in same day) $=\left(\frac{1}{6} \times \frac{1}{6}\right) \times 6=\frac{1}{6}$
(ii) $\operatorname{prob}$ (different days) $=\left(\frac{1}{6} \times \frac{5}{6}\right) \times 6=\frac{5}{6}$.
(iii) $\operatorname{prob}($ consequent days $)=\left(\frac{1}{6} \times \frac{1}{6}\right) \times 5=\frac{5}{36}$
Question 15.
In a game, the entry fee is $\square 150$. The game consists of tossing a coin 3 times. Dhana bought a ticket for entry. If one or two heads show, she gets her entry fee back. If she throws 3 heads, she receives double the entry fees. Otherwise, she will lose. Find the probability that she
(i) gets double entry fee
(ii) just gets her entry fee
(iii) loses the entry fee.
Solution:
$
\begin{aligned}
\operatorname{prob}(\text { double entry fee }) & =\operatorname{prob}(3 \mathrm{H})=\frac{1}{2 \times 2 \times 2} \\
& =\frac{1}{8}
\end{aligned}
$
prob(first gets her entry fee $)=\operatorname{prob}(1 \mathrm{H})+$
$\operatorname{prob}(2 \mathrm{H})$
$
=\frac{3}{8}+\frac{3}{8}=\frac{6}{8}=\frac{3}{4}
$
$\operatorname{prob}($ losing entry fee $)=\operatorname{prob}(0 \mathrm{H})=\frac{1}{8}$