Question

The probabilities that students will get A, B, C or D grades are 0.40, 0.35, 0.15 and 0.10 respectively. Find the probability that a student will receive
(i) not an A grade
(ii) B or C grade
(iii) at most C grade

Answer 1

Hint: To solve the above given question, we will find the probabilities of different parts according to the given question. Here, we will first find out what probability is. Probability is the ratio of favorable outcomes to the total outcomes of any event. Then, we will use the concept that the probability of X or the probability of Y is given by P(X) + P(Y).

Complete step by step solution:
Before we start solving each part of the question, we will first find out what probability is. Probability is defined as the likelihood or chance that a certain random event will occur. The probability of an event is denoted by P(E) and it is calculated by the following formula,
\[P\left( E \right)=\dfrac{\text{Favorable Outcomes}}{\text{Total Number of Outcomes}}\]
Now, we can see that there are three parts in the given question, so we will solve each part separately.
(i) not an A grade: Now the student cannot receive an A grade but he can receive B, C, or D grade. The probability that the student gets either B or C or D grade is given by:
\[\Rightarrow P\left( B,C,D \right)=P\left( B \right)+P\left( C \right)+P\left( D \right)\]
\[\Rightarrow P\left( B,C,D \right)=0.35+0.15+0.10\]
\[\Rightarrow P\left( B,C,D \right)=0.60.....\left( i \right)\]
Now, we know that, P(not A) = P(B, C, D)
\[\Rightarrow P\left( \text{not A} \right)=0.60\]
(ii) B or C grade: Now the student can either get a B grade or a C grade. The probability that the student gets either B grade or C grade is given by:
\[P\left( B,C \right)=P\left( B \right)+P\left( C \right)\]
\[\Rightarrow P\left( \text{B or C} \right)=0.35+0.15\]
\[\Rightarrow P\left( \text{B or C} \right)=0.50\]
(iii) at most C grade: Now the student can either get a C grade or a D grade. The probability that this will happen is given by,
\[P\left( \text{at most C} \right)=P\left( C \right)+P\left( D \right)\]
\[\Rightarrow P\left( \text{at most C} \right)=0.15+0.10\]
\[\Rightarrow P\left( \text{at most C} \right)=0.25\]

Note: The part (i) and (iii) of the question can also be solved as shown below.
(i) not an A grade: The probability that the student will not receive A grade will be obtained by subtracting the probability of A from 1. Thus, we have,
\[P\left( \text{not A} \right)=1-P\left( A \right)\]
\[\Rightarrow P\left( \text{not A} \right)=1-0.40\]
\[\Rightarrow P\left( \text{not A} \right)=0.60\]
(iii) At most C grade: The probability of this can be obtained by subtracting the probabilities of A and B from (i). Thus, we have,
\[\Rightarrow P\left( \text{at most C} \right)=1-\left[ P\left( A \right)+P\left( B \right) \right]\]
\[\Rightarrow P\left( \text{at most C} \right)=1-\left[ 0.40+0.35 \right]\]
\[\Rightarrow P\left( \text{at most C} \right)=1-0.75\]
\[\Rightarrow P\left( \text{at most C} \right)=0.25\]