Probability distribution of the total on two unequal fair spinners and a binomial prize condition over repeated spins
Probability distribution of the total on two unequal fair spinners and a binomial prize condition over repeated spins
12 marks
A fair spinner \(S\) has three equal sectors numbered \(1\), \(2\) and \(3\). A second fair spinner \(T\) has five equal sectors numbered \(1\), \(2\), \(3\), \(4\) and \(5\). Both spinners are spun once and the total of the two numbers obtained is recorded.
The table below shows the probability distribution of the total.
| Total | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) | \(7\) | \(8\) |
|---|---|---|---|---|---|---|---|
| Probability | \(\frac{1}{15}\) | \(p\) | \(\frac{3}{15}\) | \(q\) | \(\frac{3}{15}\) | \(r\) | \(\frac{1}{15}\) |
- Write down every pair of numbers, with the number on \(S\) first, that gives a total of \(6\).[3]
- Find the values of \(p\), \(q\) and \(r\).[3]
In a fairground game, a player spins both spinners five times. The player wins a prize if a total of \(4\) is obtained on at least two of the five turns.
- Find the probability that the player wins a prize.[6]
Solution:
Solution locked
680+ questions here come with free worked solutions. A free account saves your questions and progress; this one needs tuition enrolment.
Create a free account to practise similar questions.
Answer:(a) \((1,5),\ (2,4),\ (3,3)\) (b) \(p=\dfrac{2}{15},\ q=\dfrac15,\ r=\dfrac{2}{15}\) (c) \(\dfrac{821}{3125}\approx0.263\)
Need help? Join our JC Math tuition classes.
Learn more