Probability

Concept: Probability of an event = . A fair die has 6 faces, numbered 1, 2, 3, 4, 5, 6. Step 1: Identify the total number of possible outcomes When a fair die is thrown, the possible outcomes are \{1, 2, 3, 4, 5, 6\}. Total number of possible outcomes = 6. Step 2: Identify the favorable outcomes We want the probability of getting a number "greater than 2". The numbers on the die that are greater than 2 are \{3, 4, 5, 6\}. Number of favorable outcomes = 4. Step 3: Calculate the probability Probability (number>2) = Step 4: Simplify the fraction The fraction can be simplified by dividing both the numerator and the denominator by their greatest common divisor, which is 2. The probability of getting a number greater than 2 is . This matches option (1).

Concept: Probability of an event = . If an event is certain to happen, its probability is 1. Step 1: Determine the total number of balls in the bag Number of white balls = 5 Number of red balls = 7 Total number of balls = Number of white balls + Number of red balls Total number of balls = . This is the total number of possible outcomes when one ball is drawn. Step 2: Identify the event for which probability is to be found The event is "the ball drawn is either a white ball or a red ball." Step 3: Determine the number of favorable outcomes Since the bag only contains white balls and red balls, any ball drawn from the bag will {always} be either white or red. Number of white balls = 5 Number of red balls = 7 Number of balls that are "either white or red" = . So, the number of favorable outcomes is 12. Step 4: Calculate the probability Probability (either white or red) = This means the event is a certain event. If you draw a ball, it must be one of the colors present in the bag. This matches option (3). {Alternative approach using probability rules:} Let W be the event of drawing a white ball, and R be the event of drawing a red ball. The events W and R are mutually exclusive (a ball cannot be both white and red at the same time). The probability of "either W or R" is for mutually exclusive events. .

Concept: Probability of an event is defined as the ratio of the number of favorable outcomes to the total number of possible outcomes. In this case, we are dealing with experimental probability based on observed frequencies. Step 1: Identify the given information Let 'a' be the total number of times the coin is tossed. Let 'b' be the number of times a tail showed up. (The question uses capital A and B, but options use lowercase a and b. We will use lowercase a and b consistent with the options).
Total number of trials (tosses) =
Number of times tail appeared = Step 2: Determine the number of times head appeared A coin toss has two possible outcomes: head or tail. If the coin was tossed 'a' times and 'b' of those were tails, then the remaining tosses must have resulted in heads. Number of times head appeared = Total tosses - Number of tails Number of heads = Step 3: Calculate the probability of getting one head The question asks for "The probability of getting one head". This is interpreted as the experimental probability of a single toss resulting in a head, based on the observed data. Probability of an event = In this context:
Favorable outcome = getting a head
Number of favorable outcomes observed = Number of heads =
Total number of trials (outcomes observed) = So, the probability of getting a head, P(Head), is: Step 4: Compare with the given options The calculated probability is . This matches option (3).

Concept: Probability = (Favorable Outcomes) / (Total Outcomes). Total outcomes for two dice = . Step 1: Define events Let D1 be the outcome of the first die, D2 for the second. M2 = Multiple of 2 = \{2, 4, 6\} (3 outcomes) M3 = Multiple of 3 = \{3, 6\} (2 outcomes) We want (D1 M2 AND D2 M3) OR (D1 M3 AND D2 M2). Step 2: List favorable pairs Case A: (D1 is M2, D2 is M3) Pairs: (2,3), (2,6), (4,3), (4,6), (6,3), (6,6). (Count = pairs) Case B: (D1 is M3, D2 is M2) Pairs: (3,2), (3,4), (3,6), (6,2), (6,4), (6,6). (Count = pairs) Step 3: Combine cases and remove overlap The outcome (6,6) is in both Case A and Case B (6 is a multiple of 2 and 6 is a multiple of 3). Total unique favorable outcomes = (Outcomes in A) + (Outcomes in B) - (Overlapping outcomes) Total = (since (6,6) is the only overlap). Total favorable outcomes = 11. Step 4: Calculate probability Probability = .

Concept: Probability = (Number of favorable outcomes) / (Total number of possible outcomes). The dice are not standard identical dice; their numberings are different. Step 1: List all possible outcomes and their total number Die 1 (D1) has faces: \{1, 2, 3, 4, 5, 6\} Die 2 (D2) has faces: \{1, 1, 2, 2, 3, 3\} When two dice are thrown, the total number of possible outcomes is . Each outcome is equally likely if we consider each face of each die distinct (e.g., D2 has two '1' faces, say 1a and 1b). Step 2: Identify favorable outcomes for getting a sum of 5 We need to find pairs (Outcome on D1, Outcome on D2) such that their sum is 5. Let's list them:
If D1 = 1, D2 needs to be 4. D2 does not have 4. (No outcome)
If D1 = 2, D2 needs to be 3. D2 has two '3' faces. So, (2, 3) and (2, 3). (2 outcomes)
If D1 = 3, D2 needs to be 2. D2 has two '2' faces. So, (3, 2) and (3, 2). (2 outcomes)
If D1 = 4, D2 needs to be 1. D2 has two '1' faces. So, (4, 1) and (4, 1). (2 outcomes)
If D1 = 5, D2 needs to be 0. D2 does not have 0. (No outcome)
If D1 = 6, D2 needs to be -1. D2 does not have -1. (No outcome) The favorable pairs (D1, D2 value) are: (2, 3) - happens in 2 ways because D2 has two '3's. (3, 2) - happens in 2 ways because D2 has two '2's. (4, 1) - happens in 2 ways because D2 has two '1's. Total number of favorable outcomes = . Step 3: Calculate the probability Total number of possible outcomes = 36. Number of favorable outcomes = 6. Probability (sum = 5) =

Concept: Probability = (Number of favorable outcomes) / (Total number of possible outcomes). Step 1: Describe the initial state of the bag Initially:
Total number of balls = 14.
Number of white balls = .
Number of non-white balls = . Step 2: Describe the state of the bag after adding more white balls 6 more white balls are added to the bag. After adding:
New number of white balls = .
New total number of balls = Initial total + Added balls = . Step 3: Set up the probability equation for drawing a white ball after adding more The probability of drawing a white ball from the modified bag is given as . Using the probability formula: P(drawing a white ball) = Step 4: Solve the equation for To solve for , we can cross-multiply or multiply both sides by 20. Multiplying both sides by 20: Subtract 6 from both sides: Step 5: Check the answer (optional) If : Initially: 4 white balls, total 14 balls. After adding 6 white balls: Number of white balls = . Total number of balls = . Probability of drawing a white ball = . This matches the given information. Therefore, the value of is 4.

Concept: Probability is calculated as the ratio of the number of favorable outcomes to the total number of possible outcomes. Probability ( ) = Step 1: List all possible outcomes (Sample Space) When two coins are tossed simultaneously, let H denote a Head and T denote a Tail. The possible outcomes are:
HH (Both heads)
HT (First coin head, second coin tail)
TH (First coin tail, second coin head)
TT (Both tails) The total number of possible outcomes is 4. Step 2: Identify the favorable outcomes for "atmost one head" "Atmost one head" means zero heads OR one head. Let's list the outcomes that satisfy this condition:
Zero heads: TT (This outcome has 0 heads, which is atmost 1 head)
One head:
HT (This outcome has 1 head)
TH (This outcome has 1 head) The favorable outcomes are TT, HT, TH. The number of favorable outcomes is 3. Alternative way to think about "atmost one head": It means we do NOT want the outcome with two heads (HH). So, favorable outcomes = Total outcomes - Outcome with two heads Favorable outcomes = \{HH, HT, TH, TT\} - \{HH\} = \{HT, TH, TT\}. Number of favorable outcomes = 3. Step 3: Calculate the probability Probability (atmost one head) = Step 4: Compare with the options The calculated probability is . This matches option (2).