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The Venn diagram in Figure 1 shows the number of students in a class who read any of 3 popular magazines A, B and C - Edexcel - A-Level Maths Statistics - Question 4 - 2010 - Paper 2
Question 4
The Venn diagram in Figure 1 shows the number of students in a class who read any of 3 popular magazines A, B and C. One of these students is selected at random. (... show full transcript
Worked Solution & Example Answer:The Venn diagram in Figure 1 shows the number of students in a class who read any of 3 popular magazines A, B and C - Edexcel - A-Level Maths Statistics - Question 4 - 2010 - Paper 2
Step 1
Show that the probability that the student reads more than one magazine is \( \frac{1}{6} \)
Answer
To determine the probability of reading more than one magazine, we first need to identify the students reading multiple magazines from the Venn diagram. The total number of students is given by:
[ 4 + 2 + 5 + 3 + 6 + 10 = 30 ]
The number of students reading more than one magazine is the sum of those in overlapping areas:
[ 2 (A \cap B) + 3 (A \cap C) + 2 (B \cap C) + 5 (A \cap B \cap C) = 12 ]
Thus, the probability becomes:
[ P(more \ than \ 1) = \frac{12}{30} = \frac{2}{5} = \frac{1}{6} ]
Step 2
Find the probability that the student reads A or B (or both)
Answer
To find this probability, we use the formula for the union of two sets:
[ P(A \cup B) = P(A) + P(B) - P(A \cap B) ]
Calculating each term:
[ P(A) = \frac{4 + 2 + 5}{30} = \frac{11}{30} ]
[ P(B) = \frac{3 + 2 + 5}{30} = \frac{10}{30} ]
[ P(A \cap B) = \frac{2}{30} ]
Thus, substituting into the equation:
[ P(A \cup B) = \frac{11}{30} + \frac{10}{30} - \frac{2}{30} = \frac{19}{30} ]
Step 3
Step 4
Find the probability that the student reads C given that they read at least one magazine
Answer
To find this conditional probability, we use:
[ P(C | \text{at least one}) = \frac{P(C)}{P(\text{at least one})} ]
Here,[ P(C) = \frac{6}{30} ] (students reading C),
and ( P(\text{at least one}) = 1 - P(\text{none}) = 1 - \frac{10}{30} = \frac{20}{30} ).
Thus, substituting gives:
[ P(C | \text{at least one}) = \frac{6/30}{20/30} = \frac{6}{20} = \frac{3}{10} ]
Step 5
Determine whether or not reading magazine B and reading magazine C are statistically independent
Answer
To check for statistical independence:
We compute ( P(B) \times P(C) ) and compare it with ( P(B \cap C) ).
Calculating: [ P(B) = \frac{10}{30}, \quad P(C) = \frac{6}{30}, \quad P(B \cap C) = \frac{5}{30} ]
So, [ P(B) \times P(C) = \frac{10}{30} \times \frac{6}{30} = \frac{60}{900} = \frac{2}{30} ]
However, ( P(B \cap C) = \frac{5}{30} ) implies that: [ P(B \cap C) eq P(B) \times P(C) ]
Thus, B and C are not statistically independent.