Examples of Probability

  • If 8 red balls,7 blue balls 6 green balls.find the probabilty of picking one ball that is red or green?

    Probability can be defined as chances of occurrence of an event. Probability always lie between 0 and 1. Only in the ideal cases, it can be zero or one. If Probability of happening of an event is higher, it assures that chances of occurrence of that event is also high.
    Probability of an event X is written as P (X).  Complement of an event X is the event (not X) that is in the event of X not occurring.
    Probability in this case will be given as P (not X) = 1 – P (X).
    Joint probability of two events X and Y can be given as:
    P (X ∩ Y).
    This type of probability shows that both events X and Y happen simultaneously. This probability is known as Intersection probability. If two events are independent then intersection probability is given as:
    P (X and Y) = P (X∩ Y) = P (X) P (Y).
    If either X or Y or both occur at same time, then it is known as the Union of the events X and Y and is given mathematically as P (X U Y) and if events X and Y are mutually exclusive then probability of occurrence will be given as:
    P (X or Y) = P (X U Y) = P (X) + P (Y).
    Law of probability is called addition probability in which probability of X or Y equals to addition of probability of X and probability of Y and minus the probability of X and Y from this addition.
    Mathematically,
    P(X or Y) = P (X) + P (Y) – P (X ∩Y) where P (X ∩ Y) = 0.
     
    If 8 red balls,7 blue balls 6 green balls, find the probabilty of picking one ball that is red or green?
    Let P(X) = Probability of drawing the red ball & let P (Y) be Probability of drawing the green ball. Then Total outcomes are 17. So, P (X) = 8 / 17 & P(Y) = 6/17. Therefore P (X or Y) = 8/17 + 6/17 = 14/17.
     

    The probability of the students studying in Engineering is 40, and in medical is 50, and there are 20 students who study both, find the total number of students in the class?

    For solving this type of problem, we need to have good knowledge of Addition Theorem of Probability. According to addition theorem, if we have two Mutually Exclusive Events then we can find the Union of all the event by the formula given below:
    P(A ∪B) = P(A) +P(B)- P(A∩B)
    If we talk about our question we have  number of Students who like Engineering  so the probability will be P(A) = 30 and the number of  students who like medical are  40 so P(B) = 40 and the number of students who like both Engineering and medical  are 12 so P(A∩B) = 12 , our task is to find the number of students in the class which can be find easily by determining P(A ∪B), so by the additional formula,
    P(A ∪B) = 40 + 50 – 20,
    P(A ∪B) = 70, So the total number of students in the class is 70.

    The probability of students studying math is 30, of science is 40, and there are 12 students who study both, find the total number of student in the class?

    For solving this type of problem, we need to have good knowledge of Addition Theorem of Probability. According to addition theorem, if we have two Mutually Exclusive Events then we can find the Union of the entire event by the formula given below:
    P (A ∪B) = P (A) +P (B) - P (A∩B)
    If we talk about our question we have number of Students who like Math so probability will be P(A) = 30 and the number of  students who like Science are  40 so P(B) = 40 and the number of students who like both math and science are 12 so P(A∩B) = 12 , our task is to find the number of students in the class that we can find by finding P(A ∪B), so by the additional formula,
    P (A ∪B) = 40 + 30 – 12
    P (A ∪B) = 58
    So the total number of student in the class is 58.

    In a class there are 50 students, twenty students like playing cricket and ten students like playing football. Find the probability of student who like both football and cricket?

    For solving this type of problem, we need to have good knowledge of Addition Theorem of Probability. According to addition theorem, if we have two Mutually Exclusive Events then we can find the Union of the entire event by the formula given below:
    P (A ∪B) = P (A) +P (B) - P (A∩B)
    Here P (A) is the probability of first event and P (B) is probability of second event , P(A∩B) is the probability  of Intersection of both the event and P(A ∪B) is probability of union of both the event.
    If we talk about our question we have number of students who like cricket so probability will be P (A) = 20 and the number of  students who like football are twenty so P(B) = 10 and the number of students in the class is 50 so P(A ∪B) = 50. Now, if we see our formula only one thing is unknown that is P(A∩B), so by putting all the values we can  find the value of P(A∩B).
    50 = 20 + 10 - P (A∩B)
    Now we will take P (A∩B) to right side so there will be a change in sign,
    P (A∩B) = 30 – 50
    P (A∩B) = -20
    There is no meaning of minus sign here, so the players who play both football and cricket are 20.

    What do you mean by union of sets, explain with an example?

    Union of Sets, represented as A U B. Word Union means adding together. When we have the two Sets A and B and the resultant Set is A U B which has all the elements of set A and set B. It is read as A Union B. Let A = 1, 2, 3, 4 and B = 4, 5, 6 then we can say that
    A U B = 1, 2, 3, 4, 5, 6 which includes all the elements of set A and set B.

    Four persons are chosen from a group of 3 men, 2 women and 4 children. Find the probability of getting four persons, where exactly two persons are children?

    We use following steps for Probability of getting four persons, where exactly two persons are children -
    Step 1: First of all, we find out Sample Space of given situation means here we find out how many persons are there-
    3 + 2 + 4 = 9,
    So, there are 9 persons.
    Step 2: Now we calculate how many cases we have to retrieve four persons from there. So, total 9C4 cases because from 9 persons, we have to calculate 4 persons.
    Therefore, there are total 9C4 cases.
    Step 3: Now we find out how many ways is there for getting exactly two children from 4 persons-
    4C2 are the total number of ways for evaluating an exactly 2 children from 4 person means
    4C2.
    So, there are 4C2 ways for getting an exactly 2 children from 4 persons.
    Step 4: Now we find out how many ways is there for getting other 2 person from there-
    5C2 are the total number of ways for evaluating other 2 persons from there means
    5C2.
    So, there are 5C2 ways for getting other two persons from there.
    Step 5: Now we calculate probability of total number of favorable cases by using multiplication theorem on probability-
    Total number of favorable cases ,
    = number of ways to getting exactly 2 children * number of ways to getting other 2 person
    4C2 * 5C2
    So, total number of favorable cases is 4C2 * 5C2.
    Now we calculate probability of getting four person, where exactly two children are there
    = total favorable case / total sample space
    = (4C2 * 5C2) / 9C4,
    = 6 * 10 / 126,
    = 10 / 21,
    Therefore, probability of getting four person, where exactly two children is 10 / 21.

    Six cards are drawn at random from a pack of 52 cards. What is the probability that 3 cards are black and 3 cards are red?

    We use following steps for getting the Probability of 3 red cards and 3 black cards-
    Step 1: First of all, we find out Sample Space of given situation means here we find out how many cards are stored in a pack-
    There are 52 cards are stored in a pack.
    Step 2: Now we calculate how many cases we have to retrieve from given pack, there are 6 cards we have to retrieve in one time. So, there are total 52C6cases because in a pack, there are 52 cards and we have to retrieve 6 random cards from given pack.
    Therefore, there are 52C6 total cases.
    Step 3: Now we find out how many ways are there for getting 3 red cards from that pack-
    26C3 are the total number of ways for evaluating 3 red cards from that pack because there are 26 red cards.
    So, there are 26C3 ways for getting 3 red cards from that pack.
    Step 4: Now we find out how many ways are there for getting 3 blue cards from that pack-
    26C3 are the total number of ways for evaluating 3 blue cards from that pack because there are 26 blue cards.
    So, there are 26C3 ways for getting 3 blue cards from that pack.
    Step 5: Now we calculate probability of total number of favorable cases by using multiplication theorem on probability-
    Total number of favorable cases,
    = number of ways to getting 3 red card * number of ways to getting 3 blue card.
    26C3 * 26C3,
    So, total number of favorable cases is 26C3 * 26C3.
    Now we calculate probability that two balls drawn are red and white
    = total favorable case / total sample space,
    26C3 * 26C3 / 52C6,
    = 13000 / 39151,
    Therefore, probability of getting 6 random cards, which have 3 red cards and 3 black cards are
    13000 / 39151.

    A bag contains 3 red, 6 white and 7 blue balls. What is the probability that two balls drawn are red and white?

    We use following steps for Probability of red and white balls -
    Step 1: First of all, we find out Sample Space of given situation means here we find out how many balls are stored in a bag -
    3 + 6 + 7 = 16,
    So, there are 16 balls are stored in a bag.
    Step 2: Now we how many cases we have to retrieve from given bag, there are 2 balls we have to retrieve in one time. So, total case are-
    16C2 = (16 * 15) / (2 * 1),
           = 120.                                                                                                                
    Therefore, there are 120 total cases.
    Step 3: Now we find out how many ways is there for getting red ball in that bag-
    3C1 are the total number of ways for evaluating a red ball in that bag means,
    3C1 = 3,
    So, there are 3 ways for getting a red ball from that bag.
    Step 4: Now we find out how many ways is there for getting blue ball in that bag-
    7C1 are the total number of ways for evaluating a red ball in that bag means.
    7C1 = 7,
    So, there are 7 ways for getting a red ball from that bag.
    Step 5 : Now we calculate probability of total number of favorable cases by using multiplication theorem on probability-
    Total number of favorable cases
    = number of ways to getting red ball * number of ways to getting blue ball
    = 3 * 7
    = 21
    So, total number of favorable cases is 21.
    Now we calculate probability that two balls drawn are red and white
    = total favorable case / total sample space
    = 21 / 120
    = 7 / 40
    Therefore, probability of two balls drawn are red and white is 7 / 40.

    A card is drawn from a pack of 52 card, and drawn card is red card, now find the probability that the card drawn is greater than 7 and less than 11?

    We use following steps to find the Probability that the card drawn is greater than 7 and less than 11-
    Step 1: First of all, we assume event -
    Let e1 be the event for getting card greater than 7 and less than 11 and e2 is the event of getting red card. Step 2: Now we need to find the probability of e1 when e2 is occurred, so we need to find p (e1 / e2) -
    A pack of card contains 26 red cards and 26 black cards and we need so collect the black cards,
    So p (e1) = 26,
    Now we need to find the black cards greater than 7 and less than 11.
    As we know that in 26 black cards there will be 2 pairs of cards between 7 to 11. Now if we calculate then there will be 4 + 4 = 8 cards.
    So p (e2 ∩e1) = 8
    Now we will put that in the given formula we will get
    P (e1 / e2) = 8 / 26,
    P (e1 / e2) = 4 / 13,
    This is the required probability for the given event. So, probability that the card drawn is greater than 7 and less than 11 is p (e1 / e2) = 4 / 13.

    If probability of getting Wednesday and that a student is absent is 0.07 and there are 5 school days in a week, the probability that it is Wednesday is 0.7. Find that what is the probability that a student is absent given that day is Wednesday?

    We use following steps for finding the probability of getting the Wednesday -
    Step 1: First of all, we find out Conditional Probability means here given that the given day is Wednesday and given probability of getting Wednesday on that day is 0.7.
    Therefore, probability of getting Wednesday on that day is 0.7.
    Or P (Wednesday) = 0.7
    Step 2: Now we calculate probability of getting the day is Wednesday and that a student is absent. Here given that probability of getting the day is Wednesday and that a student is absent is 0.07. So, probability of getting the day is Wednesday and that a student is absent is 0.07 or
    P (Wednesday and absent) = 0.07,
    Step 3: Now we apply conditional probability theorem on given problem -
    P (absent | Wednesday) = P (Wednesday and absent) / P (Wednesday),
    = 0.07 / 0.7,
    = 0.10.
    So, the probability that a student is absent given that today is Wednesday is 0.10 and there are 10% chances for getting the probability that a student is absent given that day is Wednesday.

    If P(B) = 2/3 and P(A∩B)= 2 / 7 then find P(A/B)?

    Here we can easily see from the question that the Probability of occurrence of an event ‘B’ is 2 / 3.
    So P (B) = 2 / 3
    And the Intersection probability of event ‘A’ and ‘B’ is
    P (A∩B) = 2 / 7,
    Now according to the property of probability if two events ‘A’ and ‘B’ occurs then,
    Probability of ‘A’ when the Event ‘B’ is already occurred
    P (A / B) = P (A∩B) / P (B),
    = (2 / 7) / (2 / 3),
    = (6 / 14) = 3 / 7.
    So the probability of ‘A’ when the even ‘B’ is already occurred is 3 / 7.

    If P (B) = 1/5 and P (A∩B) = 1 / 8 then find P (A/B)?

    Here we can easily see from the question that the Probability of occurrence of an event ‘B’ is 1/ 5.
    So P (B) = 1 / 5,
    And the Intersection probability of event ‘A’ and ‘B’ is
    P (A∩B) = 1 / 8,
    Now according to the property of probability if two events ‘A’ and ‘B’ occurs then,
    Probability of ‘A’ when the Event ‘B’ has already occurred,
    P (A / B) = P (A∩B) / P (B),
    = (1/ 8) / (1/ 5),
    = ( 1 / 40).
    So the probability of ‘A’ when the even ‘B’ is already occurred is 1 / 40.

    If P(Y) = 1/ 4 and P(X∩Y)= 1 / 3 then find P(X/Y) ?

    Here we can easily see from the question that the Probability of occurrence an event ‘Y’ is 1 / 4.
    So P(Y) = 1/ 4,
    And the Intersection probability of event ‘X’ and ‘Y’ is
    P (X ∩ Y) = 1 / 3,
    Now according to the property of probability if two events ‘X’ and ‘Y’ occur then,
    Probability of ‘X’ when the Event ‘Y’ has already occurred,
    P(X / Y) = P (X∩Y) / P(Y),
    = (1/ 3) / (1/ 4),
    = (1 / 12).

    If P(B) = 1/3 and P(A∩B)= 1 / 5 then find P(A/B)?

    Here we can easily see from the question that the Probability of occurrence an event B is 1 / 3
    So P (B) = ( 1 / 3)
    And the Intersection probability of event ‘A’ and ‘B’ is
    P (A∩B) = (1 / 5),
    Now according to the property of probability if two events ‘A’ and ‘B’ occurs then,
    Probability of ‘A’ when the Event ‘B’ has already occurred
    P (A / B) = P (A∩B) / P (B),
    = (1/ 5) / (1/ 3),
    = (3 / 5).
    So the probability of ‘A’ when the event ‘B’ has already occurred is 3 / 5.

    In an Exam, the probability that students will pass and fail is 0.083. The probability that students will pass is 0.65. Then what is the probability that students will fail?

    The Probability of students, who will pass and fail,
    P (pass and fail) = .083,
    Probability of students who will pass is
    P (pass) = .65,
    Now the probability of student who will fail
    P (passes / fails) = P (pass and fail) / P (pass)
    = 0.083 / 0.65 = 0.127 = 12 .7%.
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