In a circle of n different persons, the total number of arrangements possible = (n – 1)!
Total number of arrangements = n(S) = (15 – 1)! = 14 !
Taking three persons as a unit, total persons = 13 (in 4 units)
Therefore no. of ways for these 13 persons to around the circular table = (13 – 1)! = 12!
In any given unit, 3 particular person can sit in 3!. Hence total number of ways that any three person can sit =
n(E) = 12! X 3!
Therefore P (E) = probability of three persons sitting together = n(E) / n(S) = 12! X 3 ! / 14!
= 12!x3x2 / 14x13x12! = 6/14×13 = 3/91
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The probabilities that a student will receive an A, B, C or D grade are 0.4, 0.3 , 0.2 and 0.1 respectively. Find the probability that a student will receive Atleast B grade.
P(atleast B) = P( B or A) = P(B) + P(A) = (0.3) + (0.4) = 0.7
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A brother and a sister appear for an interview against two vacant posts in an office. The probability of the brother’s selection is 1/5 and that of the sister’s selection is 1/3. What is the probability that only one of them is selected?
A leap year has 52 weeks and two days
Total number of cases = 7
Number of favourable cases = 1
i.e., {Friday, Saturday}
Required Probability = 1/7
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8 couples (husband and wife) attend a dance show “Nach Baliye’ in a popular TV channel ; A lucky draw in which 4 persons picked up for a prize is held, then the probability that there is atleast one couple will be selected is :
X can take 7 values.
To get |X|+2) take X={−1,0,1}
=> P(|X|<2) = Favourable CasesTotal Cases = 3/7
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I forgot the last digit of a 7-digit telephone number. If 1 randomly dial the final 3 digits after correctly dialing the first four, then what is the chance of dialing the correct number?
It is given that last three digits are randomly dialled. then each of the digit can be selected out of 10 digits in 10 ways.
Hence required probability =(110)31103 = 1/1000
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An urn contains 4 white 6 black and 8 red balls . If 3 balls are drawn one by one without replacement, find the probability of getting all white balls
Total cases of checking in the hotels = 4 x 4 x 4 = 64 ways.
Cases when 3 men are checking in different hotels = 4×3×2 = 24 ways.
Required probability =24/64 = 3/8
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Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house is :
One person can select one house out of 3= 3C13C1 ways =3.
Hence, three persons can select one house out of 3 in 3 x 3 x 3 =9.
Therefore, probability that all thre apply for the same house is 1/9
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Two brother X and Y appeared for an exam. The probability of selection of X is 1/7 and that of B is 2/9. Find the probability that both of them are selected.
Let E = event that the ball drawn is neither red nor green
= event that the ball drawn is blue.
n(E) = 7.
P(E) = n(E)/n(S) = 7/21 = 1/3.
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A bag contains 4 red and 3 black balls. A second bag contains 2 red and 4 black balls. One bag is selected at random. From the selected bag, one ball is drawn. Find the probability that the ball drawn is red.
A basket contains 10 apples and 20 oranges out of which 3 apples and 5 oranges are defective. If we choose two fruits at random, what is the probability that either both are oranges or both are non defective?
In a race where 12 cars are running, the chance that car XX will win is 1/6, that YY will win is 1/10 and that ZZ will win is 1/8. Assuming that a dead heat is impossible. Find the chance that one of them will win
A bag contains 7 green and 5 black balls. Three balls are drawn one after the other. The probability of all three balls being green, if the balls drawn are not replaced will be:
A box contains 100 balls, numbered from 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability that the sum of the three numbers on the balls selected from the box will be odd?
Total types Of French suit = 4(hearts,spades, diamonds,clubs)
Prob. = Conditional case /total case
For this conditional case of selecting all four same suit,it can be either hearts or spades or clubs or diamonds
So Total no. Of conditional case= no. of case of selecting all four cards of hearts suit +no.of case of selecting all four cards of spades suit +no.of case of selecting all four cards of clubs suit + no. of case of selecting all four cards of diamonds suit.
= 13C4+13C4+13C4+ 13C4
=4*(13C4)=2860
Total no. of case= 52C4=270725
So, probability of getting all four cards of the same suit=2860/270725
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Derek throws three dice in a special game. If he knows that he needs 15 or higher in this throw to win, then find the chance of his winning the game.
