[4 marks sum]

Question 14 Marks

Ten cards with numbers 1 to 10 written on them are placed in a bag. A card is chosen from the bag at random. Determine the probability of choosing:
(i) 7
(ii) 9 or 10
(iii) a number greater than 4
(iv) a number less than 6

Answer

Total Number of outcomes $=10$
i.e. $1,2,3,4,5,6,7,8,9,10$
(i) $P($ of getting a number 7$)=\frac{1}{10}$
(ii) P (of getting 9 or 10$)=\frac{2}{10}=\frac{1}{5}$
(iii) Numbers greater than 4 are 5, 6, 7, 8, 9 and $10=6$
P (of getting number greater than 4)
$ =\frac{6}{10}=\frac{3}{5} $
(iv) Numbers less than 6 are 1, 2, 3, 4, $5=5$
P (of getting a number less than 6)
$ =\frac{5}{10}=\frac{1}{2} $

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Question 24 Marks

A bag contains 4 white and 6 black balls,- all of the same shape and same size. A ball is drawn from the bag without looking into the bag. Find the probability that the ball drawn is:
(i) a black ball
(ii) a white ball
(iii) not a black ball

Answer

Number of white balls $=4$
Number of black balls $=6$
Number of total balls or possible event $=6+4=10$ balls
(i) Probability (a black ball)
Number of black balls $=6$
Number of total balls $=10$
$\therefore$ Probability $=\frac{6}{10}=\frac{3}{5}$
(ii) P (a white ball)
Number of white balls $=4$
Number of total balls $=10$
$\therefore$ Probability $=\frac{4}{10}=\frac{2}{5}$
(iii) P (not a black ball)
$=\frac{\text { Number of white balls }}{\text { Number of total balls }}=\frac{4}{10}=\frac{2}{5}$

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Question 34 Marks

Two coins are tossed simultaneously 100 times and the outcomes are as given below:

Outcomes	Two heads (H, H)	Exactly one head (H T or T H)	No head (T T)
No. of times	21	55	24

If the same pair of coins is tossed again at random, find the probability of getting:
(i) two heads
(ii) exactly one head
(iii) no head.

Answer

(i) Here, the total number of trials $=100$ times
Number of heads got $(H, H)=21$
$ \therefore \mathrm{P}(\mathrm{E})=\frac{\text { Number of trials in which two heads occurs }}{\text { Total number of trials }}=\frac{21}{100} $
(ii) Total number of trials $=100$ times
Number of extractly one heads $=55$
$ \therefore P(E)=\frac{55}{100}=\frac{11}{20} $
(iii) Total number of trials $=100$ times
Number of heads $=24$
$\therefore$ Probability $=\frac{24}{100}=\frac{16}{25}$

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Question 44 Marks

The following table shows number of males and number of females of a small locality in different age groups.

Age in years	10-20	21-50	Above 50
Male	8	12	6
female	6	10	4

If one of the persons, from this locality, is picked at random, what is the probability that
(a) the person picked is a male?
(b) the person picked is a female?
(c) the person picked is a female aged 21-50?
(d) the person is a male with age up to 50 years?

Answer

$\because$ Total number of persons $=$ Number of males + Number of females $=26+20=46$
(a) An event when the person picked is male $=8+12+6=26$
$\therefore$ Required Probability $=\frac{26}{46}=\frac{13}{23}$
(b) An event when the person picked is female $=6+10+4$
$\therefore$ Reqired Probability $=\frac{20}{46}=\frac{10}{23}$
(c) An event when the person picked is a female aged $21-50=10$
$\therefore$ Required Probaility $=\frac{10}{46}=\frac{5}{23}$
(d) An event when the person picked is a male aged up to 50 years $=20$
$\therefore$ Required probability $=\frac{20}{46}=\frac{10}{23}$

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Question 54 Marks

A box contains 3 yellow, 4 green, and 8 blue tickets. A ticket is chosen at random. Find the probability that the ticket is:
(i) yellow
(ii) green
(iii) blue
(iv) red
(v) not yellow

Answer

Number of yellow tickets $=3$
Number of green tickets $=4$
Number of blue tickets $=3+4+8=15$
(i) P (getting a yellow ticket) $=\frac{3}{15}=\frac{1}{5}$
(ii) P (getting a green ticket) $=\frac{4}{15}$
(iii) $P$ (getting a blue ticket) $=\frac{8}{15}$
(iv) Since, Basket contains yellow, green and blue tickets only.
$\therefore$ Number or red tickets $=0$
$\therefore P$ (getting an red ticket) $=\frac{0}{15}=0$
(v) Total number of green and blue tickets $=4+8=12$ tickets
$\mathrm{P}($ not getting yellow ticket $)=\mathrm{P}($ getting either green or blue ticket $)=\frac{12}{15}=\frac{4}{5}$
OR
$P($ not getting a yellow ticket $)=1-\frac{1}{5}=\frac{5-1}{5}=\frac{4}{5}$

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Question 64 Marks

A Ticket is randomly selected from a basket containing 3 green, 4 yellow, and 5 blue tickets. Determine the probability of getting:
(i) a green ticket
(ii) a green or yellow ticket.
(iii) an orange ticket.

Answer

Number of green tickets $=3$
Number of yellow tickets $=4$
Number of blue tickets $=5$
Total Number of tickets $=3+4+5=12$
(i) $\mathrm{P}$ (getting a green tickets) $=\frac{3}{12}=\frac{1}{4}$
(ii) Total Number of green and yellow tickets $=3+4=7$ tickets
$P($ getting a green or yellow ticket $)=\frac{7}{12}$
(iii) Since, Basket contains green, yellow and blue tickets only.
$\therefore$ Number or orange tickets $=0$
$\therefore P($ getting an orange ticket $)=\frac{0}{12}=0$

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Question 74 Marks

Five Students A, B, C, D, and E are competing in a long-distance race. Each student’s probability of winning the race is given below:
A → 20%, B → 22%, C → 7%, D → 15% and E → 36%
(i) Who is most likely to win the race?
(ii) Who is least likely to win the race?
(iii) Find the sum of probabilities given.
(iv) Find the probability that either A or D will win the race.
(v) Let S be the event that B will win the race.
(a) Find P(S)
(b) State, in words, the complementary event S’.
(c) Find P(S’)

Answer

Given probabilities of five students $A, B, C, D$, and $E$ such as
$P(A)=20 \%, P(B)=22 \%, P(C)=7 \%, P(D)=15 \%$, and $P(E)=36 \%$
(i) The mostly chance of winning the race is of Student $\mathrm{E}$........ $[\because P(E)=36 \%$ maximun $]$
(ii) The least chances of winning the race is of Student $C$......... $[\because P(C)=7 \%$ minimum $]$
(iii) The sum of the probabilities
$ \begin{aligned} & =P(A)+P(B)+P(C)+P(D)+P(E) \\ & =20 \%+22 \%+7 \%+15 \%+36 \% \\ & =100 \% \end{aligned} $
(iv) Favourable outcomes that either A or D will win $=20 \%+15 \%=35 \%$
$P\left(\right.$ either $A$ or $D$ will win) $=\frac{35}{100}=\frac{7}{20}$
(v) (a) Favourable outcomes that B will win $=22 \%$
$ P(S)=\frac{22}{100}=\frac{11}{50} $
(b) $\mathrm{S}^{\prime}=\mathrm{B}$ will not win the race.
(c) $P\left(S^{\prime}\right)=1-P(S)$
$ =1-\frac{11}{50}=\frac{50-11}{50}=\frac{39}{50} $

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MATHS [4 marks sum]

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