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A bag contains 64 coloured beads - Edexcel - A-Level Maths Statistics - Question 4 - 2018 - Paper 1
Question 4
A bag contains 64 coloured beads. There are r red beads, y yellow beads and 1 green bead and r + y + 1 = 64 Two beads are selected at random, one at a time without ... show full transcript
Worked Solution & Example Answer:A bag contains 64 coloured beads - Edexcel - A-Level Maths Statistics - Question 4 - 2018 - Paper 1
Step 1
Find the probability that the green bead is one of the beads selected.
Answer
To find the probability that the green bead is selected, we start with the following information:
- Total beads = 64
- The composition of the beads comprises ( r ) red beads, ( y ) yellow beads, and 1 green bead, such that ( r + y + 1 = 64 ).
The probability of selecting the green bead in a two-bead selection without replacement can be calculated using the formula for probabilities:
[ P(G) = P(G, R) + P(G, Y) ]
Where ( P(G, R) ) is the probability that the first bead is green and the second is red and ( P(G, Y) ) is the probability that the first bead is green and the second is yellow.
Calculating these:
- The probability that the green bead is selected first: ( P(G) = \frac{1}{64} )
- If the first bead is green, the remaining beads are 63 (( r+y )), with red and yellow beads.
- The probability of selecting a red bead next would be ( P(R|G) = \frac{r}{63} ) and for a yellow bead ( P(Y|G) = \frac{y}{63} ).
Thus, the total probability is:
[ P(G) = P(G, R) + P(G, Y) = \frac{1}{64} \cdot \frac{r}{63} + \frac{1}{64} \cdot \frac{y}{63} ]
Simplifying gives:
[ P(G) = \frac{1}{64} \cdot \frac{r+y}{63} = \frac{1}{64} \cdot \frac{64-1}{63} = \frac{1}{64} \cdot \frac{63}{63} = \frac{1}{64} \cdot 1 = \frac{1}{64} ]
Step 2
Show that r satisfies the equation r^2 - r - 240 = 0
Answer
We are given the probability that both beads are red is ( \frac{5}{84} ). This can be expressed in terms of ( r ). The probability of selecting two red beads without replacement is given by:
[ P(R_1, R_2) = \frac{r}{64} \cdot \frac{r-1}{63} = \frac{r(r-1)}{4032} ]
Setting this equal to ( \frac{5}{84} ), we solve:
[ \frac{r(r-1)}{4032} = \frac{5}{84} ]
Cross-multiply to get:
[ 84r(r-1) = 5 \cdot 4032 ]
[ 84r^2 - 84r - 20160 = 0 ]
Dividing through by 84 simplifies to:
[ r^2 - r - 240 = 0 ]
Step 3
Hence show that the only possible value of r is 16.
Answer
To find the possible values of ( r ), we can use the quadratic formula:
[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ]
For our equation ( r^2 - r - 240 = 0 ), we have( a = 1, b = -1, c = -240 ).
Calculating the discriminant:
[ b^2 - 4ac = (-1)^2 - 4\cdot 1 \cdot (-240) = 1 + 960 = 961 ]
Now apply the quadratic formula:
[ r = \frac{1 \pm \sqrt{961}}{2} = \frac{1 \pm 31}{2} ]
This gives us two solutions:
[ r = \frac{32}{2} = 16 \quad \text{and} \quad r = \frac{-30}{2} = -15 ]
Since the number of beads cannot be negative, the only possible value of ( r ) is 16.
Step 4
Given that at least one of the beads is red, find the probability that they are both red.
Answer
Let ( A ) be the event that at least one bead is red. We need to find ( P(R_1 \cap R_2 | A) ). Using conditional probability, we have:
[ P(R_1 \cap R_2 | A) = \frac{P(R_1 \cap R_2)}{P(A)} ]
From a previous step, we found ( P(R_1 \cap R_2) = \frac{5}{84} ).
To find ( P(A) ), we can calculate ( 1 - P(NR) ), the probability that no beads selected are red. The probability of selecting a non-red (yellow or green) on the first and second draws under the conditions when ( r = 16 ) is:
[ P(NR) = \frac{48}{64} \cdot \frac{47}{63} = \frac{3}{4} \cdot \frac{47}{63} = \frac{141}{252} ]
Thus,
[ P(A) = 1 - P(NR) = 1 - \frac{141}{252} = \frac{111}{252} = \frac{37}{84} ]
Now substituting into the conditional probability formula:
[ P(R_1 \cap R_2 | A) = \frac{\frac{5}{84}}{\frac{37}{84}} = \frac{5}{37} ]