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You might like to draw your own tree diagrams for these solutions.
Several of you suggested that having the same number of the same colour ribbons would make the game fair but this is not the case. Take, for example, one that many of you chose, which was four red and four blue ribbons. What happens?
\begin{eqnarray}\mbox{P (two the same colour)}&=& \mbox{P (Red and Red)}+\mbox {P (Blue and Blue)}\\ &=& \mbox{P(RR)}+\mbox{P(BB)}\\ &=&4/8 \times3/7 + 4/8 \times3/7 = 24/56 \end{eqnarray}
Laura and David of Cannock Chase High School showed that the game is not fair with 4 ribbons of one colour and two of the other
The game is not fair because:
2/6 \times1/5 = 2/30
To solve the problem, I used combinations. The formula of combinations, i.e numbers of groups of k objects taken from n objects, where the order does not matter is: {n\choose{r}} = \frac{n!}{r!(n - r)!}
1 | 3 | 3-1=2 |
3 | 6 | 6-3=3 |
6 | 10 | 10-6=4 |
10 | 15 | 15-10=5 |
15 | 21 | 21-15=6 |
21 | 28 | 28-21=7 |
Paris from IES Täby in Sweden sent in a full solution which shows algebraically the patterns that Andrei noticed.
Click here to see Paris' solution.
A man went to Monte Carlo to try and make his fortune. Is his strategy a winning one?
Two bags contain different numbers of red and blue marbles. A marble is removed from one of the bags. The marble is blue. What is the probability that it was removed from bag A?
You and I play a game involving successive throws of a fair coin. Suppose I pick HH and you pick TH. The coin is thrown repeatedly until we see either two heads in a row (I win) or a tail followed by a head (you win). What is the probability that you win?