From a group of 10 persons consisting of 5 lawyers, 3 doctors and 2 engineers, four persons are selected at random. The probability that selection contains one of each category is

Solution :

n(S) = \(^{10}C_4\) = 210

n(E)= \(^5C_2 \times ^3C_1 \times ^2C_1\) + \(^5C_1 \times ^3C_2 \times ^2C_1\) + \(^5C_1 \times ^3C_1 \times ^2C_2\) = 105

\(\therefore\) P(E) = \(105\over 210\) = \(1\over 2\)

Similar Questions

How many different words can be formed by jumbling the letters in the word ‘MISSISSIPPI’ in which no two S are adjacent ?

From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on the shelf so that the dictionary is always in the middle. Then, the number of such arrangements is

There are 10 points in a plane, out of these 6 are collinear. If N is the number of triangles formed by joining these points, then

Let \(T_n\) be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If \(T_{n+1}\) – \(T_n\) = 10, then the value of n is

The set S = {1,2,3,…..,12} is to be partitioned into three sets A, B and C of equal size. Thus, \(A\cup B\cup C\) = S \(A\cap B\) = \(B\cap C\) = \(A\cap C\) = \(\phi\) The number of ways to partition S is

Leave a Comment

Your email address will not be published.