The probabilities of a student getting 1st, 2nd, and 3rd divisions in an examination are [tex]\frac{1}{10}[/tex], [tex]\frac{3}{5}[/tex], and [tex]\frac{1}{4}[/tex] respectively. What is the probability that the student fails in the examination?

A) [tex]\frac{11}{60}[/tex]
B) [tex]\frac{1}{2}[/tex]
C) [tex]\frac{3}{20}[/tex]
D) [tex]\frac{49}{60}[/tex]

The probability of a student failing an examination, given the probabilities for passing in different divisions, is calculated by summing these probabilities and subtracting from 1. The provided options don't include the correct answer calculated as 9/20, indicating an error in the question or options.

Explanation:

The student's question is asking us to find the probability that a student fails an examination given the probabilities of achieving certain divisions in the exam. To answer this, we need to utilize our understanding of probabilities and how they add up to the total probability of an event. The probability that a student fails can be determined by summing up the probabilities of all the scenarios in which the student could pass and then subtracting that sum from 1 (since the total probability always sums up to 1).

The probabilities given for passing are:

1/2 division: 1/10
3 divisions: 3/5
1/4 division: 1/4

So, the total probability of passing would be 1/10 + 3/5 + 1/4. After adding these up and simplifying, we get a total passing probability of 11/20.

The probability the student fails would therefore be 1 - 11/20, which equals 9/20.

Since none of the given options (a: 11/60, b: 1/2, c: 3/20, d: 49/60) matches our calculation of 9/20, there must be an error in the question or in the provided options. The correct probability of failure, based on the information given, is 9/20.