RBSE Solutions Class 11 Maths Chapter 14 Probability Exercise 14.2

Rajasthan Board RBSE Class 11 Maths Chapter 14 Probability Ex 14.2

Question 1. A dice is thrown. Find the probability that number appear on dice is greater than 4.
Answer: When a single die is thrown, the total possible outcomes are the numbers 1, 2, 3, 4, 5, and 6. So, the total number of possible results, denoted as \( n(S) \), is 6. We want to find the probability that the number appearing on the die is greater than 4. The numbers greater than 4 are 5 and 6. Therefore, the number of favorable outcomes, denoted as \( n(E) \), is 2. The probability of an event is calculated by dividing the number of favorable outcomes by the total number of possible outcomes. So, the required probability is \( P(E) = \frac { n(E) }{ n(S) } = \frac { 2 }{ 6 } = \frac { 1 }{ 3 } \). This means there is a one-third chance of getting a number larger than 4.
In simple words: When you roll a dice, there are 6 possible numbers. Only two of these (5 and 6) are bigger than 4. So, the chance of getting a number bigger than 4 is 2 out of 6, which simplifies to 1 out of 3.

🎯 Exam Tip: Always list out the sample space (all possible outcomes) and the event space (favorable outcomes) clearly to avoid errors in counting.

Question 2. A coin tossed two times. Find the probability that every time tail appear.
Answer: When a coin is tossed two times, the total possible outcomes are: Head-Head (HH), Head-Tail (HT), Tail-Head (TH), and Tail-Tail (TT). The total number of outcomes, \( n(S) \), is 4. We are looking for the probability that every time a tail appears. The only outcome where both tosses result in tails is Tail-Tail (TT). So, the number of favorable outcomes, \( n(E) \), is 1. The probability is calculated as \( P(E) = \frac { n(E) }{ n(S) } = \frac { 1 }{ 4 } \). This shows that it's not a very likely event, but it is possible.
In simple words: If you flip a coin two times, there are four ways it can land. Only one of those ways is "tail" both times. So, the chance is 1 out of 4.

🎯 Exam Tip: When dealing with multiple events (like tossing a coin multiple times), systematically list all possible outcomes to ensure accuracy.

Question 3. From the natural numbers 1 to 17. One number is randomly selected. Find the probability that number is prime.
Answer: We are selecting one number from the natural numbers 1 to 17. So, the total number of possible outcomes, \( n(S) \), is 17. We need to find the probability that the selected number is a prime number. Prime numbers are whole numbers greater than 1 that have only two divisors: 1 and themselves. The prime numbers between 1 and 17 are 2, 3, 5, 7, 11, 13, and 17. The number 1 is not considered prime. Thus, the number of favorable outcomes, \( n(E) \), is 7. The required probability is \( P(E) = \frac { n(E) }{ n(S) } = \frac { 7 }{ 17 } \). Knowing the definition of prime numbers is key here.
In simple words: Out of the numbers from 1 to 17, there are 7 numbers that are prime (they can only be divided by 1 and themselves). So, the chance of picking a prime number is 7 out of 17.

🎯 Exam Tip: Remember that 1 is not a prime number. Always be careful to list all primes accurately within the given range.

Question 4. A coin is tossed three times. find the probability that alternatively head of tail appear.
Answer: When a coin is tossed three times, the total possible outcomes are 8. These outcomes form the sample space \( S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\} \). So, the total number of outcomes, \( n(S) \), is 8. We want to find the probability that head and tail appear alternatively. This means the sequence must be HTH (Head-Tail-Head) or THT (Tail-Head-Tail). The number of favorable outcomes, \( n(E) \), is 2. The required probability is \( P(E) = \frac { n(E) }{ n(S) } = \frac { 2 }{ 8 } = \frac { 1 }{ 4 } \). This shows that alternating patterns are less common than other combinations.
In simple words: If you flip a coin three times, there are 8 possible results. Only two of these results (HTH and THT) show heads and tails changing turns. So, the chance is 2 out of 8, which is 1 out of 4.

🎯 Exam Tip: For problems with multiple coin tosses, drawing a tree diagram can help ensure all possible outcomes are correctly identified.

Question 5. If two dices are thrown simultaneously, then find the probability that number appear are doublet or 2.
Answer: When two dice are thrown simultaneously, the total number of possible outcomes is \( 6 \times 6 = 36 \). Let \( S \) be the sample space, so \( n(S) = 36 \). Let \( E_1 \) be the event of getting a doublet (both dice show the same number). The outcomes for \( E_1 \) are \{(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)\}. So, \( n(E_1) = 6 \). The probability of getting a doublet is \( P(E_1) = \frac { n(E_1) }{ n(S) } = \frac { 6 }{ 36 } = \frac { 1 }{ 6 } \). Let \( E_2 \) be another event, and from the solution steps, it is used to represent an event that is mutually exclusive to \( E_1 \) and has 4 favorable outcomes. So, \( n(E_2) = 4 \). The probability of this event is \( P(E_2) = \frac { n(E_2) }{ n(S) } = \frac { 4 }{ 36 } = \frac { 1 }{ 9 } \). Since \( E_1 \) and \( E_2 \) are considered mutually exclusive events (they cannot happen at the same time), the probability of either \( E_1 \) or \( E_2 \) happening is the sum of their individual probabilities. The required probability is \( P(E_1 \text{ or } E_2) = P(E_1 \cup E_2) = P(E_1) + P(E_2) = \frac { 1 }{ 6 } + \frac { 1 }{ 9 } \). To add these fractions, we find a common denominator, which is 18. \( P(E_1 \cup E_2) = \frac { 3 }{ 18 } + \frac { 2 }{ 18 } = \frac { 3+2 }{ 18 } = \frac { 5 }{ 18 } \). This calculation assumes the event '2' refers to a set of 4 outcomes that are distinct from doublets.
In simple words: When two dice are rolled, there are 36 possible results. Getting a doublet (like two 1s or two 2s) has a chance of 6 out of 36, or 1 out of 6. If another event (which is not a doublet but related to the number 2 in some way) has a chance of 4 out of 36, or 1 out of 9, and these two events can't happen at the same time, then the chance of either of them happening is found by adding their chances together, which makes it 5 out of 18.

🎯 Exam Tip: When a question involves "or", consider whether the events are mutually exclusive. If they are, simply add their probabilities. If not, use the formula \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \).

Question 6. Find the probability that there are 52 Sundays in a normal (not leap) year.
Answer: A normal year (not a leap year) has 365 days. We know that there are 52 full weeks in a year, because \( 365 \div 7 = 52 \) with a remainder of 1. This means a normal year always has 52 Sundays, 52 Mondays, and so on, plus one extra day. This single remaining day can be any one of the seven days of the week: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, or Sunday. The probability that this extra day is a Sunday is \( \frac { 1 }{ 7 } \). For a year to have *exactly* 52 Sundays, this remaining day must *not* be a Sunday. So, the probability that there are 52 Sundays is \( 1 - P(\text{remaining day is Sunday}) = 1 - \frac { 1 }{ 7 } = \frac { 6 }{ 7 } \). This is a common probability problem involving calendars.
In simple words: A normal year has 365 days, which means 52 full weeks and one extra day. For there to be exactly 52 Sundays, that extra day must not be a Sunday. Since there are 7 possibilities for that extra day, and 6 of them are not Sunday, the chance of having exactly 52 Sundays is 6 out of 7.

🎯 Exam Tip: Clearly distinguish between a normal year and a leap year, as the number of extra days changes the probability calculation.

Question 7. A card from a deck of 52 cards is draw, find the odds in the favour of ace cad.
Answer: A standard deck of 52 cards has 4 ace cards. When a card is drawn randomly, the total number of possible outcomes, \( n(S) \), is 52. The number of favorable outcomes for drawing an ace, \( n(E) \), is 4. So, the probability of drawing an ace, \( P(E) \), is \( \frac { 4 }{ 52 } = \frac { 1 }{ 13 } \). The odds in favor of an event are calculated as the ratio of the probability of the event happening to the probability of the event not happening. Odds in favor \( = \frac { P(E) }{ 1 - P(E) } \). The probability of not drawing an ace is \( 1 - \frac { 1 }{ 13 } = \frac { 13 - 1 }{ 13 } = \frac { 12 }{ 13 } \). Therefore, the odds in favor of drawing an ace are \( \frac { \frac { 1 }{ 13 } }{ \frac { 12 }{ 13 } } = \frac { 1 }{ 12 } \). This can be written as 1:12. This means for every 1 favorable outcome, there are 12 unfavorable ones.
In simple words: There are 4 aces in a deck of 52 cards, so the chance of picking an ace is 1 out of 13. The odds in favor mean comparing the chance of picking an ace to the chance of not picking one. This is 1 chance for an ace against 12 chances for not an ace.

🎯 Exam Tip: Remember the difference between probability (a fraction) and odds (a ratio). Odds in favor are \( P(E) : (1-P(E)) \), while odds against are \( (1-P(E)) : P(E) \).

Question 8. In a class of 12 students there are 5 boys and remainings are girls. In the selection of a student find the odds against selection of girl.
Answer: In a class of 12 students, there are 5 boys. The remaining students are girls, so the number of girls is \( 12 - 5 = 7 \). The total number of students to select from, \( n(S) \), is 12. The number of favorable outcomes for selecting a girl, \( n(E_2) \), is 7. The probability of selecting a girl is \( P(E_2) = \frac { n(E_2) }{ n(S) } = \frac { 7 }{ 12 } \). The odds against selecting a girl are calculated as the ratio of the probability of *not* selecting a girl to the probability of selecting a girl. Probability of not selecting a girl is \( 1 - P(E_2) = 1 - \frac { 7 }{ 12 } = \frac { 12 - 7 }{ 12 } = \frac { 5 }{ 12 } \). This is the probability of selecting a boy. The odds against selecting a girl are \( \frac { 1 - P(E_2) }{ P(E_2) } = \frac { \frac { 5 }{ 12 } }{ \frac { 7 }{ 12 } } = \frac { 5 }{ 7 } \). This can be expressed as 5:7. This means for every 5 unfavorable outcomes (selecting a boy), there are 7 favorable outcomes (selecting a girl).
In simple words: There are 12 students, with 5 boys and 7 girls. The chance of picking a girl is 7 out of 12. The odds against picking a girl mean we compare the chance of picking a boy (5 out of 12) to the chance of picking a girl (7 out of 12). So, the odds are 5 to 7.

🎯 Exam Tip: Be careful to read whether the question asks for odds "in favor" or "against" as this flips the ratio in the final answer.

Question 9. Persons are sitting arouund a table. If two specific persons sit simultaneously. What will be the odds in this position?
Answer: Let's consider \( n \) persons sitting around a circular table. The total number of ways to arrange \( n \) distinct persons around a circular table is \( (n-1)! \). Now, let's find the number of ways two specific persons sit simultaneously (meaning next to each other). Treat these two specific persons as a single unit. Now we have \( (n-1) \) units to arrange (the pair and the remaining \( n-2 \) individuals). The number of ways to arrange these \( (n-1) \) units around a circular table is \( ((n-1)-1)! = (n-2)! \). The two specific persons within their unit can swap places in \( 2! = 2 \) ways. So, the number of favorable arrangements where the two specific persons sit together is \( 2 \times (n-2)! \). The probability \( P(A) \) that the two specific persons sit together is \( \frac { \text{Favorable arrangements} }{ \text{Total arrangements} } = \frac { 2 \times (n-2)! }{ (n-1)! } = \frac { 2 \times (n-2)! }{ (n-1) \times (n-2)! } = \frac { 2 }{ n-1 } \). The probability that the two specific persons do *not* sit together is \( 1 - P(A) = 1 - \frac { 2 }{ n-1 } = \frac { (n-1) - 2 }{ n-1 } = \frac { n-3 }{ n-1 } \). The odds against the two specific persons sitting simultaneously are given by the ratio of the probability they do not sit together to the probability they do sit together: Odds against \( = \frac { 1 - P(A) }{ P(A) } = \frac { \frac { n-3 }{ n-1 } }{ \frac { 2 }{ n-1 } } = \frac { n-3 }{ 2 } \). This ratio means for every \( n-3 \) unfavorable outcomes, there are 2 favorable ones.
In simple words: If \( n \) people sit around a table, there are many ways they can sit. If we want two special people to sit together, we can think of them as one unit. The chance they sit together is 2 divided by \( (n-1) \). The chance they don't sit together is \( (n-3) \) divided by \( (n-1) \). The "odds against" them sitting together is then the ratio of the "don't sit together" chance to the "sit together" chance, which works out to \( (n-3) \) divided by 2.

🎯 Exam Tip: For circular arrangements, remember that the total arrangements of \( n \) distinct items is \( (n-1)! \). When treating items as a unit, recalculate permutations for the units and then for items within the unit.

Question 10. There are three letters and three corresponding envelop. If each of all the letters are kept randomly in envelop, then what will be the probability that all the letters are kept in right envelops?
Answer: We have three letters and three corresponding envelopes. If each letter is put into an envelope randomly, we need to find the total number of ways these letters can be placed. This is a permutation problem. The first letter can go into any of 3 envelopes, the second into any of the remaining 2, and the third into the last 1 envelope. So, the total number of ways to place the 3 letters into 3 envelopes is \( 3 \times 2 \times 1 = 3! = 6 \). This is our sample space, \( n(S) = 6 \). We want to find the probability that all the letters are placed in their *right* envelopes. There is only one way for all three letters to be placed in their correct corresponding envelopes. So, the number of favorable outcomes, \( n(E) \), is 1. The required probability is \( P(E) = \frac { n(E) }{ n(S) } = \frac { 1 }{ 6 } \). This is a simple case of derangements where all elements are in their correct positions.
In simple words: If you have 3 letters and 3 matching envelopes, and you put them in randomly, there are 6 different ways to put them in. Only 1 of these ways has every letter in its correct envelope. So, the chance is 1 out of 6.

🎯 Exam Tip: For problems involving matching items, remember that "randomly" implies all permutations are equally likely. Always calculate total possible arrangements (permutations) and then specific favorable arrangements.

Question 11. Out of first two hundred intgers, one digit is randomly chosen.Find the probability that it will divide by 6, 8, or 24.
Answer: We are choosing one integer from the first two hundred integers, which means numbers from 1 to 200. So, the total number of possible outcomes, \( n(S) \), is 200. We need to find the probability that the chosen integer is divisible by 6, 8, or 24. Let \( A \) be the event that the number is divisible by 6. Numbers divisible by 6 from 1 to 200: \( \lfloor \frac { 200 }{ 6 } \rfloor = 33 \). So, \( n(A) = 33 \). Let \( B \) be the event that the number is divisible by 8. Numbers divisible by 8 from 1 to 200: \( \lfloor \frac { 200 }{ 8 } \rfloor = 25 \). So, \( n(B) = 25 \). Let \( C \) be the event that the number is divisible by 24. Numbers divisible by 24 from 1 to 200: \( \lfloor \frac { 200 }{ 24 } \rfloor = 8 \). So, \( n(C) = 8 \). A number divisible by both 6 and 8 must be divisible by their least common multiple (LCM), which is LCM(6, 8) = 24. So, \( n(A \cap B) = n(C) = 8 \). A number divisible by 6 and 24 is divisible by 24. So, \( n(A \cap C) = n(C) = 8 \). A number divisible by 8 and 24 is divisible by 24. So, \( n(B \cap C) = n(C) = 8 \). A number divisible by 6, 8, and 24 is divisible by 24. So, \( n(A \cap B \cap C) = n(C) = 8 \). Using the Principle of Inclusion-Exclusion for \( P(A \cup B \cup C) \): \( P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C) \) Since \( C \) includes numbers divisible by 6 and 8, the formula simplifies to: \( n(A \cup B \cup C) = n(A) + n(B) - n(A \cap B) \) because any number divisible by 24 is also divisible by 6 and 8. So, \( n(A \cup B \cup C) = n(\text{divisible by 6}) + n(\text{divisible by 8}) - n(\text{divisible by 24}) \) \( n(\text{favorable cases}) = 33 + 25 - 8 = 50 \). The required probability is \( P(\text{divisible by 6, 8, or 24}) = \frac { 50 }{ 200 } = \frac { 1 }{ 4 } \). This indicates a decent chance of picking such a number.
In simple words: Out of the first 200 numbers, we want to find the chance of picking one that can be divided evenly by 6, 8, or 24. There are 33 numbers divisible by 6, 25 by 8, and 8 by 24 (which are already counted in both 6 and 8 lists). So, we add the counts for 6 and 8, then subtract the count for 24 to avoid counting them twice. This gives us 50 such numbers. The probability is 50 out of 200, which is 1 out of 4.

🎯 Exam Tip: For "A or B or C" probability questions, always use the Principle of Inclusion-Exclusion. Remember that if one event (like divisible by 24) is a subset of others (divisible by 6 and 8), the formula simplifies.

Question 12. If three dices are thrown simultaneously. Find the probability the sum of digits obtained is more than 15.
Answer: When three dice are thrown simultaneously, each die has 6 possible outcomes. So, the total number of possible outcomes in the sample space is \( 6 \times 6 \times 6 = 6^3 = 216 \). We need to find the number of outcomes where the sum of the digits obtained is more than 15. This means the sum can be 16, 17, or 18. Let's list the combinations for each sum: 1. **Sum = 18:** * (6, 6, 6) - There is only 1 way for this combination. 2. **Sum = 17:** * (6, 6, 5) - The number 5 can be in the first, second, or third position, giving 3 permutations: (6,6,5), (6,5,6), (5,6,6). 3. **Sum = 16:** * (6, 6, 4) - 3 permutations: (6,6,4), (6,4,6), (4,6,6). * (6, 5, 5) - 3 permutations: (6,5,5), (5,6,5), (5,5,6). * Total ways for sum 16 = \( 3 + 3 = 6 \) ways. The total number of favorable outcomes (where the sum is more than 15) is the sum of ways for 18, 17, and 16: \( n(E) = 1 + 3 + 6 = 10 \). The required probability is \( P(E) = \frac { n(E) }{ n(S) } = \frac { 10 }{ 216 } \). This fraction can be simplified by dividing both the numerator and the denominator by 2: \( P(E) = \frac { 5 }{ 108 } \). This shows that obtaining such a high sum is quite rare.
In simple words: When you roll three dice, there are 216 total ways they can land. We want to find the chance that the numbers add up to more than 15 (which means 16, 17, or 18). There is 1 way to get a sum of 18 (6,6,6), 3 ways for 17 (like 6,6,5), and 6 ways for 16 (like 6,6,4 or 6,5,5). Adding these up gives 10 ways. So, the chance is 10 out of 216, which simplifies to 5 out of 108.

🎯 Exam Tip: When dealing with sums of dice, systematically list combinations starting from the highest possible sum downwards. Remember to account for all permutations of each combination.

Question 13. Letters of word ANGLE are arranged in a row randomly. Find the probability the vowels occurs together.
Answer: The word ANGLE has 5 letters. The total number of ways to arrange these 5 distinct letters in a row is \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \). This is our sample space, \( n(S) = 120 \). Now, we want to find the number of arrangements where the vowels occur together. The vowels in ANGLE are A and E. Treat these two vowels as a single block (AE). Now we are arranging 4 units: (AE), N, G, L. The number of ways to arrange these 4 units is \( 4! = 4 \times 3 \times 2 \times 1 = 24 \). Within the vowel block (AE), the vowels A and E can arrange themselves in \( 2! = 2 \times 1 = 2 \) ways (AE or EA). So, the total number of favorable arrangements where the vowels are together is \( 4! \times 2! = 24 \times 2 = 48 \). This is \( n(E) = 48 \). The required probability is \( P(E) = \frac { n(E) }{ n(S) } = \frac { 48 }{ 120 } \). This fraction can be simplified: divide by 24, \( \frac { 48 \div 24 }{ 120 \div 24 } = \frac { 2 }{ 5 } \). So, the probability is 2/5.
In simple words: The word ANGLE has 5 letters. There are 120 ways to arrange these letters. If we want the vowels (A and E) to always stay next to each other, we can think of them as one group. Then we arrange this group and the other letters, which gives 24 ways. Since A and E can also swap places (AE or EA), we multiply by 2. This makes 48 ways the vowels are together. The chance is 48 out of 120, which simplifies to 2 out of 5.

🎯 Exam Tip: When a question asks for specific letters/items to be "together", always treat them as a single block for the main arrangement, and then multiply by the internal permutations of the items within that block.

Question 14. A card is drawn from a deck of card. Find the probability that chosen card is ace, king or queen.
Answer: A standard deck of 52 cards consists of various ranks and suits. We want to find the probability that a chosen card is an ace, a king, or a queen. Let \( A \) be the event of drawing an ace. There are 4 aces in a deck. So, \( P(A) = \frac { 4 }{ 52 } = \frac { 1 }{ 13 } \). Let \( K \) be the event of drawing a king. There are 4 kings in a deck. So, \( P(K) = \frac { 4 }{ 52 } = \frac { 1 }{ 13 } \). Let \( Q \) be the event of drawing a queen. There are 4 queens in a deck. So, \( P(Q) = \frac { 4 }{ 52 } = \frac { 1 }{ 13 } \). These three events (drawing an ace, a king, or a queen) are mutually exclusive, meaning you cannot draw a card that is both an ace and a king at the same time. Therefore, the probability of drawing an ace, king, or queen is the sum of their individual probabilities: \( P(A \text{ or } K \text{ or } Q) = P(A) + P(K) + P(Q) \) \( P(A \text{ or } K \text{ or } Q) = \frac { 1 }{ 13 } + \frac { 1 }{ 13 } + \frac { 1 }{ 13 } = \frac { 1 + 1 + 1 }{ 13 } = \frac { 3 }{ 13 } \). This is a fairly straightforward calculation for mutually exclusive events.
In simple words: In a deck of 52 cards, there are 4 aces, 4 kings, and 4 queens. Since you can only pick one card at a time, picking an ace, a king, or a queen are separate things. So, we add their chances: 4 out of 52 for an ace, plus 4 out of 52 for a king, plus 4 out of 52 for a queen. This totals 12 out of 52, which simplifies to 3 out of 13.

🎯 Exam Tip: When events are mutually exclusive (cannot happen at the same time), the probability of "A or B" is simply \( P(A) + P(B) \). For drawing cards, different ranks are always mutually exclusive.

Question 15. A bag contains 6 white, 7 red and 5 black balls. Out of these 3 balls are ramdomly chosen one by one.What will be the probability that three balls are white,whereas the ball drawn is not replaced back?
Answer: The bag contains 6 white balls, 7 red balls, and 5 black balls. The total number of balls in the bag is \( 6 + 7 + 5 = 18 \). We are choosing 3 balls one by one without replacement, and we want to find the probability that all three balls drawn are white. 1. **Probability of the first ball being white (P(A)):** There are 6 white balls out of 18 total. \( P(A) = \frac { 6 }{ 18 } = \frac { 1 }{ 3 } \). 2. **Probability of the second ball being white (P(B)), given the first was white and not replaced:** After drawing one white ball, there are 5 white balls left and a total of 17 balls remaining in the bag. \( P(B) = \frac { 5 }{ 17 } \). 3. **Probability of the third ball being white (P(C)), given the first two were white and not replaced:** After drawing two white balls, there are 4 white balls left and a total of 16 balls remaining in the bag. \( P(C) = \frac { 4 }{ 16 } = \frac { 1 }{ 4 } \). The probability that all three balls drawn are white (P(A and B and C)) is the product of these individual probabilities, as the events are dependent: \( P(A \cap B \cap C) = P(A) \times P(B) \times P(C) = \frac { 1 }{ 3 } \times \frac { 5 }{ 17 } \times \frac { 1 }{ 4 } \). \( P(A \cap B \cap C) = \frac { 1 \times 5 \times 1 }{ 3 \times 17 \times 4 } = \frac { 5 }{ 204 } \). This is a fairly low probability, as expected for multiple specific draws without replacement.
In simple words: You have a bag with 18 balls, 6 of which are white. You pick three balls one after another, and you don't put them back. The chance of the first ball being white is 6 out of 18. If that happens, now there are 5 white balls left out of 17 total, so the chance of the second being white is 5 out of 17. If that also happens, there are 4 white balls left out of 16 total, making the third chance 4 out of 16. To find the chance of all three happening, you multiply these chances: 1/3 times 5/17 times 1/4, which gives 5 out of 204.

🎯 Exam Tip: For "without replacement" problems, always adjust both the number of favorable outcomes and the total number of outcomes after each draw. Dependent events require multiplying conditional probabilities.