Probability Class 9 CBSE Important Questions with Answers

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) $\dfrac12$1/2.

The sample space is $\{1,2,3,4,5,6\}${1,2,3,4,5,6}, with $6$6 equally likely outcomes. The primes are $2,3,5$2,3,5, so $P(\text{prime})=\dfrac{3}{6}=\dfrac12$P(prime)=3/6=1/2. (Remember that $1$1 is not a prime number.)

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.

Sample space $=\{HH,\ HT,\ TH,\ TT\}$={HH, HT, TH, TT}, with $4$4 equally likely outcomes. ($HT$HT and $TH$TH are different outcomes: the first coin shows heads in one and tails in the other.)

(i) Exactly one head: $HT,\ TH$HT, TH, so $P=\dfrac24=\dfrac12$P=2/4=1/2.

(ii) No head: $TT$TT only, so $P=\dfrac14$P=1/4.

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

This equals the theoretical probability $\dfrac14$1/4, 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.

(iii) Theoretically, two heads is $1$1 of the $4$4 equally likely outcomes $\{HH,HT,TH,TT\}${HH,HT,TH,TT}, so $P=\dfrac14=0.25$P=1/4=0.25. The experimental value $0.27$0.27 is close to it; with more tosses the experimental probability usually gets closer to the theoretical one.

Each die has $6$6 faces, so there are $6\times6=36$6×6=36 equally likely outcomes. The table shows the sum for each pair:

(i) Sum $7$7: $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$(1,6),(2,5),(3,4),(4,3),(5,2),(6,1), i.e. $6$6 outcomes, so $P=\dfrac{6}{36}=\dfrac16$P=6/36=1/6.

(ii) Same number (a doublet): $(1,1),\dots,(6,6)$(1,1),…,(6,6), i.e. $6$6 outcomes, so $P=\dfrac{6}{36}=\dfrac16$P=6/36=1/6.

(iii) The product is odd only when both numbers are odd: $3\times3=9$3×3=9 outcomes, so $P(\text{odd})=\dfrac{9}{36}=\dfrac14$P(odd)=9/36=1/4 and $P(\text{even})=1-\dfrac14=\dfrac34$P(even)=1-1/4=3/4.

(a) Certain: $P=1$P=1.

(b) Impossible (a die shows only $1$1 to $6$6): $P=0$P=0.

(c) $P=\dfrac12$P=1/2, in the middle of the scale (equally likely to happen or not).

(d) There are $13$13 hearts, so $P(\text{heart})=\dfrac{13}{52}=\dfrac14$P(heart)=13/52=1/4 and $P(\text{not a heart})=1-\dfrac14=\dfrac34$P(not a heart)=1-1/4=3/4.

On the scale: (b) at $0$0, (c) at $\tfrac12$1/2, (d) at $\tfrac34$3/4, (a) at $1$1. Since $\tfrac34>\tfrac12$3/4>1/2, event (d) is more likely than (c).

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.

(a) Each toss branches into H and T:

There are $2\times2\times2=8$2×2×2=8 equally likely outcomes: $HHH, HHT, HTH, HTT, THH, THT, TTH, TTT$HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.

(b) Exactly two heads: $HHT, HTH, THH$HHT, HTH, THH, so $P=\dfrac38$P=3/8.

(c) 'At least one head' is the complement of 'no head' ($TTT$TTT only): $P=1-\dfrac18=\dfrac78$P=1-1/8=7/8.

(d) All the same: $HHH$HHH or $TTT$TTT, so $P=\dfrac28=\dfrac14$P=2/8=1/4.

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.