Li has $t$ toy bricks - OCR - GCSE Maths - Question 8 - 2020 - Paper 6
Question 8
Li has $t$ toy bricks.
She only has red bricks and blue bricks.
Li picks two bricks, one after the other.
If the first brick she picks is red, the probability ... show full transcript
Worked Solution & Example Answer:Li has $t$ toy bricks - OCR - GCSE Maths - Question 8 - 2020 - Paper 6
Step 1
If the first brick she picks is red, the probability that the second brick is red is $\frac{2}{3}$
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Answer
Let the number of red bricks be r and the number of blue bricks be b.
The total number of bricks is t=r+b.
If the first brick is red, there will be r−1 red bricks and b blue bricks left.
Thus, the probability of the second brick being red after picking the first red brick is: P(R2∣R1)=t−1r−1=32
From this we can set up the equation: 3(r−1)=2(t−1)
Expanding this gives us: 3r−3=2t−2
So we have: 3r−2t=1(1)
Step 2
If the first brick she picks is blue, the probability that the second brick is red is $\frac{7}{10}$
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Answer
If the first brick is blue, there will be r red bricks and b−1 blue bricks left.
Thus, the probability of the second brick being red after picking the first blue brick is: P(R2∣B1)=t−1r=107
From this we can set up the equation: 10r=7(t−1)
Expanding this gives us: 10r=7t−7
So we have: 10r−7t=−7(2)
Step 3
Solve the simultaneous equations
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Answer
We now have the two equations:
3r−2t=1
10r−7t=−7
To eliminate r, we can multiply equation (1) by 10 and equation (2) by 3:
30r−20t=10
30r−21t=−21
Subtracting these equations gives us: −20t+21t=10+21 t=31
Now substituting t=31 back into equation (1) to find r: 3r−2(31)=1 3r−62=1 3r=63 r=21
Thus, the total number of bricks is: t=31.
