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Question 15 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 25 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 35 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 45 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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[5 marks sum] - MATHS STD 7 Questions - Vidyadip