Solution:
A number is divisible by 4 if and only if the last 2 digits of the number is divisible by 4. Thus these numbers must have the last 2 digits as 24 or 12 or 32 or 52. There are a total of \(4\times 3!=24\) such numbers. Total number of numbers formed by these digits is \(5!=120\). Thus required probability = \(\frac{24}{120}=\frac{1}{5}\).
Q-2) There are 8 floors of a house, and 5 persons enter the lift at the ground floor. Each of them are independent and can get down in any floor of the house that they want, starting from the first. Find the probability that all 5 of them leave at different floors.
Solution:
Let’s denote the sample space as S and the event that all persons get down at different floors as E. Thus the number of events in E is \(^7P_5\). While, the total number of ways the persons can get down is \(7^5\). Thus the required probability is \(\frac{^7P_5}{7^5}\).
Q-3) From a set X containing n elements, 2 subsets A and B are chosen at random. Find the probability that \(A\bigcup B=X\).
Solution:
For each \(x_i\in X\), we have 4 possibilities:
a) \(x_i\) is in both A and B
b) \(x_i\) is neither in A nor in B
c) \(x_i\) is in A but not in B
d) \(x_i\) is in B but not in A
We should not have d) in our case for the criteria to be fulfilled.
Denoting the total sample space as S, and the events favourable for \(A\bigcup B=X\), we have \(n(S)=4^n\) and \(n(E)=3^n\). Thus required probability = \(\frac{3^n}{4^n}\). Note that as \(n\) increases, this probability decreases.
Q-4) 2 teams A and B play a tournament. The first team to win \(n+1\) games wins the tournament. Probability for A winning a game is \(p\), and that of B winning is \(q\). Assuming that there are no ties, find the probability that A wins the tournament.
Solution:
Suppose that there are \(n+r+1\) games that A needs to win the tournament, and let’s say he wins the \((n+r+1)\)th game and \(n\) out of the first \((n+r)\) games. Thus the probability for A winning the tournament is \(\sum\limits_{r=0}^nC(n+r,n)p^{n+1}q^r\) – which is what we need. However, exploring these probabilities a bit further gives an interesting identity.
Similarly the probability of B winning the tournament is \(\sum\limits_{r=0}^nC(n+r,n)q^{n+1}p^r\).
Adding these 2 probabilities is likely to give us 1, thus
\(P(A)+P(B)=1=\sum\limits_{r=0}^nC(n+r,n)(q^rp^{n+1}+q^{n+1}p^r)\)Let’s substitute \(p=q=\frac{1}{2}\). That gives us
\(\sum\limits_{r=0}^nC(n+r,n)\left[ \frac{1}{2^{n+r+1}} +\frac{1}{2^{n+r+1}}\right]=1\Rightarrow \sum\limits_{r=0}^nC(n+r,n)\frac{1}{2^{n+r}}=1\).
Q-5) If 3 numbers are randomly chosen from the 1st \(2n+1\) numbers, what is the probability that these numbers will form an A.P?
Solution:
We have, according to our earlier notations, \(n(S)=C(2n+1,3)=\frac{n(4n^2-1)}{3}\). Let’s consider our triplets with common difference on a case by case basic:
\(d=1\Rightarrow (1,2,3),(2,3,4),…,(2n-1,2n,2n+1)\), thus a total of 2n-1 triplets.
\(d=2\Rightarrow (1,3,5),(2,4,6),…,(2n-3,2n-1,2n+1)\), thus a total of 2n-3 triplets.
…………
…………
\(d=n-1\Rightarrow (1,n,2n-1),(2,n+1,2n),(3,n+2,2n+1)\), thus a total of 3 triplets
\(d=n\Rightarrow (1,n+1,2n+1)\), thus a total of 1 triplet.
Thus the total number of triplets that can be chosen such that its elements are in A.P is \(1+3+5+…+2n-1=n^2\).
That gives our probability as \(\frac{n(E)}{n(S)}=\frac{3n}{4n^2-1}\).
Q-6) A speaks truth 3 times out of 4, and B 7 times out of 10; they both assert that a white ball has been drawn from a bag containing 6 balls all of different colours : find the probability that the ball is white.
Solution:
Let’s denote by \(P_1\), the event that the ball is white, and let \(P_2\) denote the event that the ball is not white. From the existing data we can conclude that \(P_1=\frac{1}{6}\) and \(P_2=\frac{5}{6}\). Also let \(p_1\) denote the event that both A and B are saying the truth, and \(p_2\) denote the event that they both lie independently, given that a white ball has been drawn. Thus, \(p_1=\frac{3}{4}\times \frac{7}{10}\) and \(p_2=\left(\frac{1}{5}\times \frac{1}{4}\right)\times \left(\frac{1}{5}\times\frac{3}{10}\right)\). From Baye’s Theorem, we have the required probability as
\(\frac{P_1p_1}{P_1p_1+p_2p_2}=\frac{35}{36}\).
Q-7) From each of two equal lines of length l, a portion is cut off at random, and removed : what is the probability that the sum of the remainders is less than l?
Solution:
Place the lines parallel to one another, and suppose that after cutting, the right-hand portions are removed. Then the question is equivalent to asking what is the chance that the sum of the right-hand portions is greater than the sum of the left-hand portions. It is clear that the first sum is equally likely to be greater or less than the second; thus the required probability is \(\frac{1}{2}\).
Q-8) If three lines are chosen at random, what is the probability that they denote the sides of a triangle?
Solution:
Of three lines one must be equal to or greater than each of the other two ; denote its length by l. Then all we know of the other two lines is that the length of each lies between 0 and l. But if each of two lines is known to be of random length between 0 and l, then, going by what we showed in the above example, the probability that their sum is greater than l is \(\frac{1}{2}\). Which basically means that the probability that they form the sides of a triangle is also \(\frac{1}{2}\).
Q-9) Three tangents are drawn at random to a given circle, what is the probability of the circle being inscribed in the triangle formed by them.
Solution:
Consider the figure below:
Draw three random lines P, (), 11, in the same plane as the circle, and draw to the circle the six tangents parallel to these lines. Now, among the 8 triangles so formed, the circle is inscribed in only 2 of them. Thus the required probability happens to be \(\frac{2}{8}=\frac{1}{4}\).
Q-10) If we make 4 throws with a pair of dice, what is the probability that we end up having doublets (Both the dices show the same number) at least twice?
Solution:
In a single throw the chance of doublets is \(\frac{6}{36}=\frac{1}{6}\), while the chance of failing to do the same is \(\frac{5}{6}\).
Now the required event follows if doublets are thrown four times, three times, or twice ; therefore the required probability is the sum of the first 3 terms of the expansion \(\left(\frac{1}{6}+\frac{5}{6} \right)^4\)