Home / IB Math Analysis & Approaches Question bank-Topic: SL 4.7 Applications SL Paper 1

IB Math Analysis & Approaches Question bank-Topic: SL 4.7 Applications SL Paper 1

Question

Jar A contains three red marbles and five green marbles. Two marbles are drawn from the jar, one after the other, without replacement.

Jar B contains six red marbles and two green marbles. A fair six-sided die is tossed. If the score is \(1\) or \(2\), a marble is drawn from jar A. Otherwise, a marble is drawn from jar B.

Find the probability that

  (i)     none of the marbles are green;

  (ii)     exactly one marble is green.

[5]
a.

Find the expected number of green marbles drawn from the jar.

[3]
b.

(i)     Write down the probability that the marble is drawn from jar B.

(ii)     Given that the marble was drawn from jar B, write down the probability that it is red.

[2]
c.

Given that the marble is red, find the probability that it was drawn from jar A.

[6]
d.
Answer/Explanation

Markscheme

(i)     attempt to find \({\rm{P(red)}} \times {\rm{P(red)}}\)     (M1)

eg   \(\frac{3}{8} \times \frac{2}{7}\) , \(\frac{3}{8} \times \frac{3}{8}\) , \(\frac{3}{8} \times \frac{2}{8}\)

\({\text{P(none green)}} = \frac{6}{{56}}\) \(\left( { = \frac{3}{{28}}} \right)\)     A1     N2

(ii)     attempt to find \({\rm{P(red)}} \times {\rm{P(green)}}\)     (M1)

eg   \(\frac{5}{8} \times \frac{3}{7}\) , \(\frac{3}{8} \times \frac{5}{8}\) , \(\frac{{15}}{{56}}\)

recognizing two ways to get one red, one green     (M1)

eg   \(2{\rm{P}}(R) \times {\rm{P}}(G)\) , \(\frac{5}{8} \times \frac{3}{7} + \frac{3}{8} \times \frac{5}{7}\) , \(\frac{3}{8} \times \frac{5}{8} \times 2\)

\({\text{P(exactly one green)}} = \frac{{30}}{{56}}\) \(\left( { = \frac{{15}}{{28}}} \right)\)     A1     N2

[5 marks]

a.

\({\text{P(both green)}} = \frac{{20}}{{56}}\) (seen anywhere)     (A1)

correct substitution into formula for \({\rm{E}}(X)\)     A1

eg   \(0 \times \frac{6}{{56}} + 1 \times \frac{{30}}{{56}} + 2 \times \frac{{20}}{{56}}\) , \(\frac{{30}}{{64}} + \frac{{50}}{{64}}\)

expected number of green marbles is \(\frac{{70}}{{56}}\) \(\left( { = \frac{5}{4}} \right)\)     A1     N2

[3 marks]

b.

(i)     \({\text{P(jar B)}} = \frac{4}{6}\) \(\left( { = \frac{2}{3}} \right)\)     A1     N1

(ii)     \({\text{P(red| jar B)}} = \frac{6}{8}\) \(\left( { = \frac{3}{4}} \right)\)     A1     N1

[2 marks]

c.

recognizing conditional probability     (M1)

eg   \({\rm{P}}(A|R)\) , \(\frac{{{\text{P(jar A and red)}}}}{{{\rm{P(red)}}}}\) , tree diagram

attempt to multiply along either branch (may be seen on diagram)     (M1)

eg   \({\text{P(jar A and red)}} = \frac{1}{3} \times \frac{3}{8}\) \(\left( { = \frac{1}{8}} \right)\)

attempt to multiply along other branch     (M1)

eg   \({\text{P(jar B and red)}} = \frac{2}{3} \times \frac{6}{8}\) \(\left( { = \frac{1}{2}} \right)\)

adding the probabilities of two mutually exclusive paths     (A1)

eg   \({\rm{P(red)}} = \frac{1}{3} \times \frac{3}{8} + \frac{2}{3} \times \frac{6}{8}\)

correct substitution

eg   \({\text{P(jar A|red)}} = \frac{{\frac{1}{3} \times \frac{3}{8}}}{{\frac{1}{3} \times \frac{3}{8} + \frac{2}{3} \times \frac{6}{8}}}\) , \(\frac{{\frac{1}{8}}}{{\frac{5}{8}}}\)     A1

\({\text{P(jar A|red)}} = \frac{1}{5}\)     A1     N3

[6 marks]

d.

Question

The following diagram shows a board which is divided into three regions \(A\), \(B\) and \(C\).

A game consists of a contestant throwing one dart at the board. The probability of hitting each region is given in the following table.

Find the probability that the dart does not hit the board.

[3]
a.

The contestant scores points as shown in the following table.

Given that the game is fair, find the value of \(q\).

[4]
b.
Answer/Explanation

Markscheme

evidence of summing probabilities to \(1\)     (M1)

eg\(\;\;\;\frac{5}{{20}} + \frac{4}{{20}} + \frac{1}{{20}} + p = 1,\;\;\;\sum { = 1} \)

correct working     (A1)

eg\(\;\;\;p = 1 – \frac{{10}}{{20}}\)

\(p = \frac{{10}}{{20}}\;\;\;\left( { = \frac{1}{2}} \right)\)     A1     N2

[3 marks]

a.

correct substitution into \({\text{E}}(X)\)     (A1)

eg\(\;\;\;\frac{4}{{20}}(q) + \frac{1}{{20}}(10) + \frac{{10}}{{20}}( – 3)\)

valid reasoning for fair game (seen anywhere, including equation)     (M1)

eg\(\;\;\;{\text{E}}(X) = 0\), points lost\( = \)points gained

correct working     (A1)

eg\(\;\;\;4q + 10 – 30 = 0,\;\;\;\frac{4}{{20}}q + \frac{{10}}{{20}} = \frac{{30}}{{20}}\)

\(q = 5\)     A1     N2

[4 marks]

Total [7 marks]

b.
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