Question

261.- In room 1 there are seven students, of which four are studying engineering and three will act, in room 2 there are 8 students, of which three are studying engineering and five are acting. A random student from room 2 is moved to room 1 and then a random student from room 1 is chosen. Determine the probability that he is an engineering student.

263.-urn 1 contains two white and two black spheres, urn 2 contains two white and three black. A random urn is selected and two spheres are extracted together in an aletoria way. What is the probability that they are the same color?

Answer 1

261:

Let E1 shows the event that an engineering student is selected from room 1 and A1 shows the event that a act student is selected from room 1.

Let E2 shows the event that an engineering student is selected from room 2 and A2 shows the event that a act student is selected from room 2.

Total number of students in room 2 : 8

Out of 8 students three are studying engineering and five are acting so

P(E2 ) = 3/8 =0.375, P(A2) = 5/8 = 0.625

After moving 1 student from room2 to room1 total number of students in room 1 is : 7+1 = 8

P(E1 | E2) = (4+1) / 8 = 0.625, P(E1 | A2) = (4) / 8 = 0.5

So the probability that an engineering student is selected will be

P(E1) = P(E1|E2)P(E2) +P(E1|A2)P(A2) = 0.625 * 0.375 + 0.5 * 0.625= 0.546875

Answer: 0.5469