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Probability - Simple problems on single events

Grade 10ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Random Experiment and Outcomes: A random experiment is a process where the result cannot be predicted with certainty. For example, tossing a fair coin is an experiment where the possible outcomes are Heads (HH) or Tails (TT). Visualize a coin spinning in the air; until it lands, the outcome is unknown, but we know it must be one of these two results.

Sample Space (SS): The set of all possible outcomes of a random experiment is called the sample space. For a single throw of a six-sided die, the sample space is S={1,2,3,4,5,6}S = \{1, 2, 3, 4, 5, 6\}, which can be visualized as the six distinct faces of the cube labeled with dots.

Event (EE): An event is a subset of the sample space consisting of one or more outcomes. For instance, in a die roll, the event 'getting an even number' corresponds to the set E={2,4,6}E = \{2, 4, 6\}. Visualize this as highlighting only the specific outcomes you are looking for within the entire sample space.

Equally Likely Outcomes: Outcomes are said to be equally likely if none of them is expected to occur in preference to others. In a well-shuffled deck of 5252 cards, every card has the same chance of being picked. Visualize a balanced scale where every individual outcome carries the same 'weight' or chance.

Probability Scale: The probability of an event is always a real number between 00 and 11 inclusive. Visualize a number line where 00 represents an 'Impossible Event' (like rolling a 77 on a standard die) and 11 represents a 'Sure Event' (like rolling a number less than 77 on a standard die).

Complementary Events: For every event EE, there is an event 'not EE' (denoted as EE' or Eˉ\bar{E}) which occurs when EE does not occur. Visualize a Venn diagram where a rectangular box is the whole sample space and a circle inside represents EE; the region outside the circle but inside the box is EE'. The sum of their probabilities is always 11.

Deck of Cards Composition: A standard deck contains 5252 cards divided into 44 suits: Hearts, Diamonds, Spades, and Clubs. Visualize 1313 cards in each suit (A,2,3,4,5,6,7,8,9,10,J,Q,KA, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K). Hearts and Diamonds are red (2626 cards), while Spades and Clubs are black (2626 cards). There are 1212 'face cards' (Jacks, Queens, Kings) in total.

Theoretical Probability: This is the ratio of the number of outcomes favorable to an event to the total number of possible outcomes in a sample space. Visualize this as a fraction where the 'target' outcomes are the numerator and the 'total' possibilities are the denominator.

📐Formulae

P(E)=Number of outcomes favorable to ETotal number of possible outcomes=n(E)n(S)P(E) = \frac{\text{Number of outcomes favorable to } E}{\text{Total number of possible outcomes}} = \frac{n(E)}{n(S)}

0P(E)10 \le P(E) \le 1

P(E)+P(Eˉ)=1P(E) + P(\bar{E}) = 1 or P(not E)=1P(E)P(\text{not } E) = 1 - P(E)

P(ϕ)=0P(\phi) = 0 (Probability of an impossible event)

P(S)=1P(S) = 1 (Probability of a sure event)

💡Examples

Problem 1:

A bag contains 55 red balls, 88 white balls, and 44 green balls. A ball is drawn at random from the bag. Find the probability that the ball drawn is (i) white, (ii) not green.

Solution:

Step 1: Find the total number of outcomes n(S)n(S). Total balls =5+8+4=17= 5 + 8 + 4 = 17. So, n(S)=17n(S) = 17. \nStep 2: For (i), let WW be the event of drawing a white ball. The number of favorable outcomes n(W)=8n(W) = 8. \nStep 3: Calculate P(W)=n(W)n(S)=817P(W) = \frac{n(W)}{n(S)} = \frac{8}{17}. \nStep 4: For (ii), let GG be the event of drawing a green ball. n(G)=4n(G) = 4. The event 'not green' is Gˉ\bar{G}. \nStep 5: Calculate P(Gˉ)=1P(G)=1417=17417=1317P(\bar{G}) = 1 - P(G) = 1 - \frac{4}{17} = \frac{17-4}{17} = \frac{13}{17}. (Alternatively, favorable outcomes for 'not green' are red or white balls: 5+8=135 + 8 = 13)

Explanation:

This problem uses the basic definition of probability and the concept of complementary events. We first identify the total count (sample space) and then the specific counts for the target events.

Problem 2:

A card is drawn from a well-shuffled pack of 5252 playing cards. Find the probability that the card drawn is (i) a face card, (ii) a red king.

Solution:

Step 1: Total number of possible outcomes n(S)=52n(S) = 52. \nStep 2: For (i), face cards are Jacks, Queens, and Kings. There are 33 face cards in each of the 44 suits. Total face cards n(F)=3×4=12n(F) = 3 \times 4 = 12. \nStep 3: P(Face Card)=1252P(\text{Face Card}) = \frac{12}{52}. Simplifying by dividing both by 44, we get 313\frac{3}{13}. \nStep 4: For (ii), identify red kings. The red suits are Hearts and Diamonds. Each has 11 king. Total red kings n(KR)=2n(K_R) = 2. \nStep 5: P(Red King)=252=126P(\text{Red King}) = \frac{2}{52} = \frac{1}{26}.

Explanation:

To solve card problems, you must know the composition of the deck. We identify the subset of cards that satisfy the condition (face cards or red kings) and divide by the total number of cards.