The Mathematics of Maybe: Introduction to Probability — CBSE Class 9 Maths Important Questions

Answer: (a) $0\le P(E)\le1$0≤ P(E)≤1.

A probability is a fraction of favourable to total trials, so it can never be negative nor exceed $1$1.

Answer: (a) $0.56$0.56.

$P(\text{head})=\dfrac{\text{number of heads}}{\text{total tosses}}=\dfrac{56}{100}=0.56$P(head)=number of heads/total tosses=56/100=0.56.

Answer: (b) $0.85$0.85.

$P(\text{six})=\dfrac{30}{200}=0.15$P(six)=30/200=0.15, so $P(\text{not six})=1-0.15=0.85$P(not six)=1-0.15=0.85.

Answer: (a) $0.30$0.30.

Blue draws $=500-220-130=150$=500-220-130=150. So $P(\text{blue})=\dfrac{150}{500}=0.30$P(blue)=150/500=0.30.

Answer: (a) Both A and R are true and R is the correct explanation of A. With $0$0 favourable trials, $P(E)=\dfrac{0}{\text{total}}=0$P(E)=0/total=0.

Total tosses $=150$=150.

(i) $P(\text{tail})=\dfrac{81}{150}=\dfrac{27}{50}=0.54$P(tail)=81/150=27/50=0.54.

(ii) Heads $=150-81=69$=150-81=69, so $P(\text{head})=\dfrac{69}{150}=\dfrac{23}{50}=0.46$P(head)=69/150=23/50=0.46.

$P(\text{spade})=\dfrac{65}{260}=\dfrac{1}{4}=0.25$P(spade)=65/260=1/4=0.25.

This equals the theoretical probability $\dfrac14$14, so the experimental result matches the expected value closely.

Total trials $=500$=500.

(i) $P(\text{two heads})=\dfrac{135}{500}=0.27$P(two heads)=135/500=0.27.

(ii) 'At least one head' means two heads or one head $=135+255=390$=135+255=390 times.

$P(\text{at least one head})=\dfrac{390}{500}=0.78$P(at least one head)=390/500=0.78.

Total throws $=300$=300.

(i) $P(4)=\dfrac{60}{300}=\dfrac{1}{5}=0.2$P(4)=60/300=1/5=0.2.

(ii) Even outcomes are $2,4,6$2,4,6: frequency $=55+60+52=167$=55+60+52=167.

$P(\text{even})=\dfrac{167}{300}\approx0.557$P(even)=167/3000.557.

Families with $2$2 girls: $P=\dfrac{\text{number}}{400}=0.35\Rightarrow\text{number}=0.35\times400=140$P=number/400=0.35=0.35×400=140 families.

Families with no girl $=60$=60, so $P(\text{no girl})=\dfrac{60}{400}=0.15$P(no girl)=60/400=0.15.

$P(\text{at least one girl})=1-P(\text{no girl})=1-0.15=0.85$P(at least one girl)=1-P(no girl)=1-0.15=0.85.

Total balls $=80$=80.

(i) Boundaries $=4$=4s and $6$6s $=8+4=12$=8+4=12 balls. $P(\text{boundary})=\dfrac{12}{80}=\dfrac{3}{20}=0.15$P(boundary)=12/80=3/20=0.15.

(ii) No run (a dot ball) $=28$=28 balls. $P(\text{no run})=\dfrac{28}{80}=\dfrac{7}{20}=0.35$P(no run)=28/80=7/20=0.35.

(iii) Scored at least $1$1 run $=80-28=52$=80-28=52 balls. $P(\text{at least one run})=\dfrac{52}{80}=\dfrac{13}{20}=0.65$P(at least one run)=52/80=13/20=0.65.

Check: $P(\text{no run})+P(\text{at least one run})=0.35+0.65=1$P(no run)+P(at least one run)=0.35+0.65=1, as it must, since these two events cover all outcomes.

Total bulbs $=50+120+180+150=500$=50+120+180+150=500.

(i) $P(\text{less than }1)=\dfrac{50}{500}=0.1$P(less than 1)=50/500=0.1.

(ii) At least $2$2 years $=$= '2 to 3' $+$+ 'more than 3' $=180+150=330$=180+150=330. $P=\dfrac{330}{500}=0.66$P=330/500=0.66.

(iii) Less than $3$3 years $=50+120+180=350$=50+120+180=350. $P=\dfrac{350}{500}=0.7$P=350/500=0.7.

(iv) Estimated number lasting at least $2$2 years $=0.66\times2000=1320$=0.66×2000=1320 bulbs.

Total students $=40$=40.

(i) $P(\text{cricket})=\dfrac{16}{40}=\dfrac{2}{5}=0.4$P(cricket)=16/40=2/5=0.4.

(ii) Football or hockey $=10+6=16$=10+6=16. $P=\dfrac{16}{40}=0.4$P=16/40=0.4.

(iii) Not badminton $=40-8=32$=40-8=32. $P=\dfrac{32}{40}=\dfrac{4}{5}=0.8$P=32/40=4/5=0.8.

(iv) $\dfrac{16}{40}+\dfrac{10}{40}+\dfrac{6}{40}+\dfrac{8}{40}=\dfrac{40}{40}=1$16/40+10/40+6/40+8/40=40/40=1. The probabilities sum to $1$1, as required.