Probability — Exercise 14.1

Complementary events cover the sample space without overlap. An impossible event has no favourable outcomes; a certain event includes every outcome. All elementary outcomes together make the whole sample space.

(i) Car starting and not starting need not be equally likely: the probabilities depend on the car’s condition. (ii) Making or missing a basketball shot depends on skill and conditions. (iii) The standard textbook model treats a blind random choice between true and false as equally likely, giving 1/2 for each; a knowledgeable answerer need not have a 1/2 success rate. (iv) The elementary textbook model assumes boy/girl outcomes are equally likely; this is an idealised model rather than an exact demographic claim.

For an unbiased coin, heads and tails have the same probability, 1/2. Assign one team to each face before tossing. Each then has the same chance.

Probabilities lie in the closed interval from 0 to 1. The other values are 2/3, 0.15 and 0.7, all inside it.

Use P(not E)=1−P(E)=1−0.05.

Orange candy has zero favourable outcomes. Every candy is lemon, so drawing lemon is certain.

Use the complement of the event as stated in the question: 1−0.992.

There are 8 equally likely balls. Red has 3 favourable outcomes; not red has the other 5.

Total=5+8+4=17. Red count=5; white count=8; not-green count=5+8=13. Divide each by 17.

The total number of coins is 180; use coin counts rather than their monetary values. P(50p)=100/180=5/9. There are 180−10=170 coins that are not ₹5.

There are 13 fish and 5 favourable choices, assuming each fish is equally likely to be selected.

The eight numbers are equally likely. (i) Only 8 qualifies: 1 outcome. (ii) Odd numbers are 1,3,5,7: 4 outcomes. (iii) Greater than 2: 3,4,5,6,7,8, giving 6 outcomes. (iv) All eight numbers are below 9.

The sample space is {1,2,3,4,5,6}. Primes are {2,3,5}; numbers strictly between 2 and 6 are {3,4,5}; odds are {1,3,5}. Each set contains 3 of the 6 outcomes.

There are 2 red kings; 12 face cards (J,Q,K in four suits); 6 red face cards; 1 jack of hearts; 13 spades; and 1 queen of diamonds. Divide by 52. Aces are not face cards.

Initially exactly one of five cards is a queen, so P(Q)=1/5. After the queen is removed, four cards remain including one ace and no queen. The second draw uses the updated four-card sample space.

Total=144. Good outcomes=132. Thus 132/144 simplifies by dividing numerator and denominator by 12.

(i) P(defective)=4/20=1/5. (ii) Initially 16 are good. Removing one good bulb leaves 15 good among 19 total.

Two-digit numbers run from 10 through 90: 90−10+1=81. Positive squares are 1,4,9,16,25,36,49,64,81: 9. Multiples of 5 are 5,10,…,90: 18. Divide by 90.

Count faces, not distinct letter names. A appears on two of six faces and D on one.

This geometric model assumes the landing point is uniformly distributed over the rectangle and ignores the physical die’s size. Radius=1/2 m; circle area=π/4 m². Rectangle area=6 m². Probability equals favourable area divided by total area.

Good pens=144−20=124. She buys a good pen, so P(buy)=124/144=31/36. She refuses a defective pen: 20/144=5/36. The two sum to 1.

Two independent fair dice produce 6×6=36 equally likely ordered pairs. Counts for sums 2,3,4,5,6,7,8,9,10,11,12 are respectively 1,2,3,4,5,6,5,4,3,2,1. For example sum 2 has only (1,1), but sum 7 has (1,6),(2,5),(3,4),(4,3),(5,2),(6,1).

Eight equally likely ordered sequences are HHH,HHT,HTH,HTT,THH,THT,TTH,TTT. Two win, six lose.

There are 36 ordered outcomes. To avoid 5, each roll has 5 permitted faces, giving 5×5=25 outcomes. The event of at least one 5 is the complement, with 36−25=11 outcomes.

(i) Incorrect. HH,HT,TH,TT are the four equally likely ordered outcomes. Two heads has probability 1/4, two tails 1/4, and one of each 2/4. The three categories are not equally likely. (ii) Correct for a fair die: odd faces {1,3,5} and even faces {2,4,6} each contain three of six equally likely elementary outcomes.