All questions of Probability for Electrical Engineering (EE) Exam

Total number of outcomes possible when a die is rolled = 6 (∵ any one face out of the 6 faces)

Total number of balls = 2 + 3 + 2 = 7

► Let S be the sample space.

► Let E = Event of drawing 2 balls, none of them is blue.

Total count of numbers, n(S) = 10

Prime numbers in the given range are 2,3,5 and 7
Hence, total count of prime numbers in the given range, n(E) = 4

Total number of balls = 4 + 5 + 6 = 15

Let S be the sample space.

Let E = Event of drawing 3 balls, all of them are yellow.

[∵ ⁿC_r = ⁿC_(n-r). So ⁵C₃ = ⁵C₂. Applying this for the ease of calculation]

 

Let S be the sample space.

Let E = Event of selecting 1 girl and 2 boys

15 boys and 10 girls are there in a class. We need to select 2 boys from 15 boys and 1 girl from 10 girls

Number of ways in which this can be done: 
¹⁵C₂ × ¹⁰C₁
Hence n(E) = ¹⁵C₂ × ¹⁰C₁

he period from July 4 to July 8, inclusive, contains 8 – 4 + 1 = 5 days, so we can rephrase the question as “What is the probability of having exactly 3 rainy days out of 5?” Since there are 2 possible outcomes for each day (R = rain or S = shine) and 5 days total, there are 2 x 2 x 2 x 2 x 2 = 32 possible scenarios for the 5 day period (RRRSS, RSRSS, SSRRR, etc…) To find the probability of having exactly three rainy days out of five, we must find the total number of scenarios containing exactly 3 R’s and 2 S’s, that is the number of possible RRRSS anagrams: = 5! / 2!3! = (5 x 4)/2 x 1 = 10 The probability then of having exactly 3 rainy days out of five is 10/32 or 5/16. Note that we were able to calculate the probability this way because the probability that any given scenario would occur was the same. This stemmed from the fact that the probability of rain = shine = 50%. Another way to solve this question would be to find the probability that one of the favorable scenarios would occur and to multiply that by the number of favorable scenarios. In this case, the probability that RRRSS (1st three days rain, last two shine) would occur is (1/2)(1/2)(1/2)(1/2)(1/2) = 1/32. There are 10 such scenarios (different anagrams of RRRSS) so the overall probability of exactly 3 rainy days out of 5 is again 10/32. This latter method works even when the likelihood of rain does not equal the likelihood of shine. 

To find the probability of forming a code with two adjacent I’s, we must find the total number of such codes and divide by the total number of possible 10-letter codes. The total number of possible 10-letter codes is equal to the total number of anagrams that can be formed using the letters ABCDEFGHII, that is 10!/2! (we divide by 2! to account for repetition of the I's).
To find the total number of 10 letter codes with two adjacent I’s, we can consider the two I’s as ONE LETTER. The reason for this is that for any given code with adjacent I’s, wherever one I is positioned, the other one must be positioned immediately next to it. For all intents and purposes, we can think of the 10 letter codes as having 9 letters (I-I is one). There are 9! ways to position 9 letters.
Probability = (# of adjacent I codes) / (# of total possible codes) = 9! ÷ (10! / 2! ) = ( 9!2! / 10! ) = (9!2! / 10(9!) ) = 1/5

The correct option is A 

 

P(E) =  (3/4 × 1/7) + (1/4 × 2/7)
P(Leap Year/ E) = (2/28) / (5/28) = 2/5 

 

Total number of cards, n(S) = 52
Total number of black cards, n(E) = 26

Total number of cards = 52

Total Number of King Cards = 4

Total Number of Diamond Cards = 13

Total Number of Cards which are both King and Diamond = 1

Here a card can be both a Diamond card and a King. Hence these are not mutually exclusive events. (Reference : mutually exclusive events) . By Addition Theorem of Probability, we have P(King or a Diamond) = P(King) + P(Diamond) – P(King and Diamond)

Every player has an equal chance of leaving at any particular time. Thus, the probability that four particular players leave the field first is equal to the probability that any other four players leave the field first. In other words, the answer to this problem is completely independent of which four players leave first.
Given the four players that leave first, there are 4! or 24 orders in which these players can leave the field - only one of which is in increasing order of uniform numbers. (For example, assume the players have the numbers 1, 2, 3, and 4. There are 24 ways to arrange these 4 numbers: 1234, 1243, 1324, 1342, 1423, 1432, . . . , etc. Only one of these arrangements is in increasing order.)
Thus, the probability that the first four players leave the field in increasing order of their uniform numbers is 1/24. 

The simplest way to approach a complex probability problem is not always the direct way. In order to solve this problem directly, we would have to calculate the probabilities of all the different ways we could get two oppositehanded, same-colored gloves in three picks. A considerably less taxing approach is to calculate the probability of NOT getting two such gloves and subtracting that number from 1 (remember that the probability of an event occurring plus the probability of it NOT occurring must equal 1).
Let's start with an assumption that the first glove we pick is blue. The hand of the first glove is not important; it could be either right or left. So our first pick is any blue. Since there are 3 pairs of blue gloves and 10 gloves total, the probability of selecting a blue glove first is 6/10.
Let's say our second pick is the same hand in blue. Since there are now 2 blue gloves of the same hand out of the 9 remaining gloves, the probability of selecting such a glove is 2/9.
Our third pick could either be the same hand in blue again or any green. Since there is now 1 blue glove of the same hand and 4 green gloves among the 8 remaining gloves, the probability of such a pick is (1 + 4)/8 or 5/8. The total probability for this scenario is the product of these three individual probabilities: 6/10 x 2/9 x 5/8 = 60/720.
We can summarize this in a chart:

We can apply the same principles to our second scenario, in which we choose blue first, then any green, then either the same-handed green or the same-handed blue:

But it is also possible to pick green first. We could pick any green, then the same-handed green, then any blue:

Or we could pick any green, then any blue, then the same-handed green or same-handed blue:


The overall probability of NOT getting two gloves of the same color and same hand is the SUM of the probabilities of these four scenarios: 60/720 + 72/720 + 24/720 + 72/720 = 228/720 = 19/60.
Therefore, the probability of getting two gloves of the same color and same hand is 1 - 19/60 = 41/60.

In order to solve this problem, we have to consider two different scenarios. In the first scenario, a woman is picked from room A and a woman is picked from room B. In the second scenario, a man is picked from room A and a woman is picked from room B. The probability that a woman is picked from room A is 10/13. If that woman is then added to room B, this means that there are 4 women and 5 men in room B (Originally there were 3 women and 5 men). So, the probability that a woman is picked from room B is 4/9.
Because we are calculating the probability of picking a woman from room A AND then from room B, we need to multiply these two probabilities: 10/13 x 4/9 = 40/117 The probability that a man is picked from room A is 3/13. If that man is then added to room B, this means that there are 3 women and 6 men in room B. So, the probability that a woman is picked from room B is 3/9.
Again, we multiply thse two probabilities: 3/13 x 3/9 = 9/117 To find the total probability that a woman will be picked from room B, we need to take both scenarios into account.
In other words, we need to consider the probability of picking a woman and a woman OR a man and a woman. In probabilities, OR means addition. If we add the two probabilities, we get: 40/117 + 9/117 = 49/117

For probability, we always want to find the number of ways the requested event could happen and divide it by the total number of ways that any event could happen.
For this complicated problem, it is easiest to use combinatorics to find our two values. First, we find the total number of outcomes for the triathlon. There are 9 competitors; three will win medals and six will not. We can use the Combinatorics Grid, a counting method that allows us to determine the number of combinations without writing out every possible combination.

Out of our 9 total places, the first three, A, B, and C, win medals, so we label these with a "Y." The final six places (D, E, F, G, H, and I) do not win medals, so we label these with an "N." We translate this into math: 9! / 3!6! = 84. So our total possible number of combinations is 84. (Remember that ! means factorial; for example, 6! = 6 × 5 × 4 × 3 × 2 × 1.)
Note that although the problem seemed to make a point of differentiating the first, second, and third places, our question asks only whether the brothers will medal, not which place they will win. This is why we don't need to worry about labeling first, second, and third place distinctly.
Now, we need to determine the number of instances when at least two brothers win a medal. Practically speaking, this means we want to add the number of instances two brothers win to the number of instances three brothers win.
Let's start with all three brothers winning medals, where B represents a brother.

Since all the brothers win medals, we can ignore the part of the counting grid that includes those who don't win medals. We have 3! / 3! = 1. That is, there is only one instance when all three brothers win medals.
Next, let's calculate the instances when exactly two brothers win medals.

Since brothers both win and don't win medals in this scenario, we need to consider both sides of the grid (i.e. the ABC side and the DEFGHI side). First, for the three who win medals, we have 3! / 2! = 3. For the six who don't win medals, we have 6! / 5! = 6. We multiply these two numbers to get our total number: 3 × 6 = 18.
Another way to consider the instances of at least two brothers medaling would be to think of simple combinations with restrictions. If you are choosing 3 people out of 9 to be winners, how many different ways are there to chose a specific set of 3 from the 9 (i.e. all the brothers)? Just one. Therefore, there is only one scenario of all three brothers medaling.
If you are choosing 3 people out of 9 to be winners, if 2 specific people of the 9 have to be a member of the winning group, how many possible groups are there? It is best to think of this as a problem of choosing 1 out of 7 (2 must be chosen). Choosing 1 out of 7 can be represented as 7! / 1!6! = 7. However, if 1 of the remaining 7 can not be a member of this group (in this case the 3rd brother) there are actually only 6 such scenarios. Since there are 3 different sets of exactly two brothers (B1B2, B1B3, B2B3), we would have to multiply this 6 by 3 to get 18 scenarios of only two brothers medaling. The brothers win at least two medals in 18 + 1 = 19 circumstances. Our total number of circumstances is 84, so our probability is 19 / 84.

If set S is the set of all prime integers between 0 and 20 then: S = {2, 3, 5, 7, 11, 13, 17, 19}
Let’s start by finding the probability that the product of the three numbers chosenis a number less than 31. To keep the product less than 31, the three numbers must be 2, 3 and 5. So, what is the probability that the three numbers chosen will be some combination of 2, 3, and 5?
Here’s the list all possible combinations of 2, 3, and 5: case A: 2, 3, 5 case B: 2, 5, 3 case C: 3, 2, 5 case D: 3, 5, 2 case E: 5, 2, 3 case F: 5, 3, 2 This makes it easy to see that when 2 is chosen first, there are two possible combinations. The same is true when 3 and 5 are chosen first. The probability of drawing a 2, AND a 3, AND a 5 in case A is calculated as follows (remember, when calculating probabilities, AND means multiply): case A: (1/8) x (1/7) x (1/6) = 1/336 The same holds for the rest of the cases. case B: (1/8) x (1/7) x (1/6) = 1/336 case C: (1/8) x (1/7) x (1/6) = 1/336 case D: (1/8) x (1/7) x (1/6) = 1/336 case E: (1/8) x (1/7) x (1/6) = 1/336 case F: (1/8) x (1/7) x (1/6) = 1/336 So, a 2, 3, and 5 could be chosen according to case A, OR case B, OR, case C, etc. The total probability of getting a 2, 3, and 5, in any order, can be calculated as follows (remember, when calculating probabilities, OR means add): (1/336) + (1/336) + (1/336) + (1/336) + (1/336) + (1/336) = 6/336
Now, let’s calculate the probability that the sum of the three numbers is odd. In order to get an odd sum in this case, 2 must NOT be one of the numbers chosen. Using the rules of odds and evens, we can see that having a 2 would give the following scenario: even + odd + odd = even So, what is the probability that the three numbers chosen are all odd? We would need an odd AND another odd, AND another odd: (7/8) x (6/7) x (5/6) = 210/336 The positive difference between the two probabilities is: (210/336) – (6/336) = (204/336) = 17/28

The simplest way to solve the problem is to recognize that the total number of gems in the bag must be a multiple of 3, since we have 2/3 diamonds and 1/3 rubies. If we had a total number that was not divisible by 3, we would not be able to divide the stones into thirds. Given this fact, we can test some multiples of 3 to see whether any fit the description in the question. The smallest number of gems we could have is 6: 4 diamonds and 2 rubies (since we need at least 2 rubies). Is the probability of selecting two of these diamonds equal to 5/12? 4/6 × 3/5 = 12/30 = 2/5. Since this does not equal 5/12, this cannot be the total number of gems. The next multiple of 3 is 9, which yields 6 diamonds and 3 rubies: 6/9 × 5/8 = 30/72 = 5/12. Since this matches the probability in the question, we know we have 6 diamonds and 3 rubies. Now we can figure out the probability of selecting two rubies: 3/9 × 2/8 = 6/72 = 1/12

Total number of outcomes possible when a coin is tossed = 2 (? Head or Tail)
Hence, total number of outcomes possible when 5 coins are tossed, n(S) = 2⁵

E = Event of getting exactly 2 heads when 5 coins are tossed
n(E) = Number of ways of getting exactly 2 heads when 5 coins are tossed = ⁵C₂

Total number of outcomes possible when a die is rolled = 6 (? any one face out of the 6 faces)

Hence, total number of outcomes possible when two dice are rolled = 6 × 6 = 36

To get a sum of 7, the following are the favourable cases.
(1, 6), (2, 5), {3, 4}, (4, 3), (5, 2), (6,1)

=> Number of ways in which we get a sum of 7 = 6

To get a sum of 11, the following are the favourable cases. (5, 6), (6, 5) => Number of ways in which we get a sum of 11 = 2

Here, clearly the events are mutually exclusive events. By Addition Theorem of Probability, we have P(a sum of 7 or a sum of 11) = P(a sum of 7) + P( a sum of 11)

Since each die has 6 possible outcomes, there are 6 × 6 = 36 different ways that Bill can roll two dice. Similarly there are 6 × 6 = 36 different ways for Jane to roll the dice. Hence, there are a total of 36 × 36 = 1296 different possible ways the game can be played.
One way to approach this problem (the hard way) is to consider, in turn, the number of ways that Bill can get each possible score, compute the number of ways that Jane can beat him for each score, and then divide by 1296. The number of ways to make each score is: 1 way to make a 2 (1 and 1), 2 ways to make a 3 (1 and 2, or 2 and 1), 3 ways to make a 4 (1 and 3, 2 and 2, 3 and 1), 4 ways to make a 5 (use similar reasoning…), 5 ways to make a 6, 6 ways to make a 7, 5 ways to make an 8, 4 ways to make a 9, 3 ways to make a 10, 2 ways to make an 11, and 1 way to make a 12. We can see that there is only 1 way for Bill to score a 2 (1 and 1). Since there are 36 total ways to roll two dice, there are 35 ways for Jane to beat Bob's 2. 
Next, there are 2 ways that Bob can score a 3 (1 and 2, 2 and 1). There are only three ways in which Jane would not beat Bob: if she scores a 2 (1 and 1), she would lose to Bob or if she scores a 3 (1 and 2, 2 and 1), she would tie Bob. Since there are 36 total ways to roll the dice, Jane has 33 ways to beat Bob.
Using similar logic, we can quickly create the following table:

Out of the 1296 possible ways the game can be played, 575 of them result in Jane winning the game. Hence, the probability the Jane will win is 575/1296 and the correct answer is C.
There is a much easier way to compute this probability. Observe that this is a “symmetric” game in that neither Bill nor Jane has an advantage over the other. That is, each has an equal change of winning. Hence, we can determine the number of ways each can win by subtracting out the ways they can tie and then dividing the remaining possibilities by 2. 
Note that for each score, the number of ways that Jane will tie Bill is equal to the number of ways that Bill can make that score (i.e., both have an equal number of ways to make a particular score). Thus, referring again to the table above, the total number of ways to ti e are: 12 + 22 + 32 + 42 +52 +62 + 52 + 42 + 32 + 22 + 12 = 146. Therefore, there are 1296 – 146 = 1150 non-ties. Since this is a symmetric problem, Jane will win 1150/2 or 575 times out of the 1296 possible games. Hence, the probability that she will win is 575/1296.

The easiest way to attack this problem is to pick some real, easy numbers as values for y and n . Let's assume there are 3 travelers (A, B, C) and 2 different destinations (1, 2). We can chart out the possibilities as follows: 

Thus there are 8 possibilities and in 2 of them all travelers end up at the same destination. Thus the probability is 2/8 or 1/4. By plugging in y = 3 and n = 2 into each answer choice, we see that only answer choice D yields a probability of 1/4. Alternatively, consider that each traveler can end up at any one of n destinations. Thus, for each traveler there are n possibilities. Therefore, for y travelers, there are possible outcomes. Additionally, the "winning" outcomes are those where all travelers end up at the same destination. Since there are n destinations there are n "winning" outcomes.
Thus, the probability = 

Total number of cards = 52

Total number of Ace cards = 4

Total number of cards, n(S) = 52

Total number of "Queen" cards, n(E) = 4

Let A = the event that John is selected and B = the event that Dani is selected.

Given that 𝑃(𝐴) = 1/3 and 𝑃(𝐵) = 1/5

We know that A is the event that A does not occur and B is the event that B does not occur

Probability that only one of them is selected

There are 2 possible outcomes on each flip: heads or tails. Since the coin is flipped three times, there are 2 × 2 × 2 = 8 total possibilities: HHH, HHT, HTH, HTT, TTT, TTH, THT, THH.
Of these 8 possibilities, how many involve exactly two heads? We can simply count these up: HHT, HTH, THH. We see that there are 3 outcomes that involve exactly two heads. Thus, the correct answer is 3/8.
Alternatively, we can draw an anagram table to calculate the number of outcomes that involve exactly 2 heads.

The top row of the anagram table represents the 3 coin flips: A, B, and C. The bottom row of the anagram table represents one possible way to achieve the desired outcome of exactly two heads. The top row of the anagram yields 3!, which must be divided by 2! since the bottom row of the anagram table contains 2 repetitions of the letter H.
There are 3!/2! = 3 different outcomes that contain exactly 2 heads.
The probability of the coin landing on heads exactly twice is (# of two-head results) ÷ (total # of outcomes) = 3/8.

Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}

Let E = event of getting at most two heads.

Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}.

Therefore  P(E) = n(E)/n(S) = 7/8.