A biased five-sided spinner

A biased five-sided spinner

Secondary 2
6 marks
TGM Original Questions

A biased spinner has the table:

Number\(1\)\(2\)\(3\)\(4\)\(5\)
Probability\(\frac{1}{4}\)\(x\)\(\frac{1}{5}\)\(\frac{1}{10}\)\(x\)
  1. Find \(x\).[2]
  2. Find the probability of
    1. 2 or 3,[1]
    2. an odd number,[1]
    3. a number not more than 4.[1]
  3. In 80 spins, find the expected number of results equal to 3 or 4.[1]

Solution:

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Answer:(a) \(x=\frac{9}{40}\) (b)(i) \(\frac{17}{40}\) (ii) \(\frac{27}{40}\) (iii) \(\frac{31}{40}\) (c) \(24\).

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