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The bag P contains 6 balls of which 3 are red and 3 are yellow - Edexcel - A-Level Maths Statistics - Question 7 - 2011 - Paper 1
Question 7
The bag P contains 6 balls of which 3 are red and 3 are yellow. The bag Q contains 7 balls of which 4 are red and 3 are yellow. A ball is drawn at random from ba... show full transcript
Worked Solution & Example Answer:The bag P contains 6 balls of which 3 are red and 3 are yellow - Edexcel - A-Level Maths Statistics - Question 7 - 2011 - Paper 1
Step 1
Complete the tree diagram shown below.
Answer
The completed tree diagram would show the probabilities of drawing red or yellow from bag P and how those affect the probabilities in bag Q. The relevant paths in the tree involve:
-
Selecting Red from Bag P (Probability = \frac{1}{2})
- Then selecting Red from Q: (Probability = \frac{4}{9})
- Then third ball drawn (from 9): ( \frac{5}{9} )
- Total =
-
Selecting Yellow from Bag P (Probability = \frac{1}{2})
- Then selecting Red from Q: (Probability = \frac{5}{9})
- Results in ( \frac{5}{9} ) as well.
Step 2
Find P(A)
Answer
To find P(A), we need the probability that the two balls drawn from bag P are of the same color. The two scenarios are both Red or both Yellow:
- Both Red: ( P(RR) = \frac{3}{6} \times \frac{2}{5} = \frac{1}{5} )
- Both Yellow: ( P(YY) = \frac{3}{6} \times \frac{2}{5} = \frac{1}{5} )
Therefore, ( P(A) = P(RR) + P(YY) = \frac{1}{5} + \frac{1}{5} = \frac{2}{5} ).
Step 3
Show that P(B) = \frac{5}{9}
Answer
To find P(B), we consider all outcomes where a red ball is drawn from bag Q. We compute their probabilities:
- If a RED ball came from bag P: ( P(RB) = \frac{1}{2} \times \frac{4}{9} = \frac{2}{9} )
- If a YELLOW ball came from bag P: ( P(YB) = \frac{1}{2} \times \frac{5}{9} = \frac{5}{18} )
Thus, the total probability:
( P(B) = P(RB) + P(YB) = \frac{2}{9} + \frac{5}{18} = \frac{5}{9}. )
Step 4
Show that P(A \cap B) = \frac{2}{9}
Answer
To find ( P(A \cap B) ) we look at the cases where A and B can occur simultaneously.
- From the scenario where both balls are red (from event A) occurring and resulting in a red ball from bag Q, we have:
( P(RR \cap R) = P(RR) * P(R | RR) = \frac{1}{5} ) - Hence ( P(A \cap B) = \frac{1}{5} + \frac{1}{45} = \frac{2}{9} ).
Step 5
Hence find P(A \cup B)
Answer
To find ( P(A \cup B) ), we can use the formula:
[ P(A \cup B) = P(A) + P(B) - P(A \cap B) ]
- Substituting the values we found:
( P(A) = \frac{2}{5}, P(B) = \frac{5}{9}, P(A \cap B) = \frac{2}{9} ) - This results in
[ P(A \cup B) = \frac{2}{5} + \frac{5}{9} - \frac{2}{9} = \frac{11}{45} ]
Thus, the required probability is ( P(A \cup B) = \frac{11}{45} ).
Step 6
Given that all three balls drawn are the same colour, find the probability that they are all red.
Answer
Firstly, calculate the probabilities of drawing three balls of the same colour:
- All Red:
- From bag P: ( P(RRR) = \frac{3}{6} \cdot \frac{2}{5} \cdot \frac{1}{4} = \frac{1}{20} )
- All Yellow:
- Similarly, ( P(YYY) = \frac{3}{6} \cdot \frac{2}{5} \cdot \frac{1}{4} = \frac{1}{20} )
- Total for the same colour: ( P(Same) = P(RRR) + P(YYY) = \frac{1}{20} + \frac{1}{20} = \frac{1}{10} )
Now, the probability that all are red given that all three drawn are the same colour is:
[ P(RRR | Same) = \frac{P(RRR)}{P(Same)} = \frac{\frac{1}{20}}{\frac{1}{10}} = \frac{1}{2} ]