John flipped a coin five times and obtained five successive heads. Is it conclusive that the coin is biased? Explain your answer clearly. A two-digit number is formed by drawing a random card from Bag A followed by another random card from Bag B. Bag A contains three cards labelled '1', '2' and '3'. Bag B contains three cards labelled '4', '5' and '7'. For example, drawing a '1' and '5' from Bag A and Bag B respectively will form the number '15'. Find, as a fraction in its simplest form, the probability that the two-digit number formed is Complete the possibility diagram below. odd, Answer ... a perfect square, Answer ... a prime number, Answer ... a factor of 135. Answer ...

1. Recognise that five heads is possible for a fair coin

For a fair coin, getting heads on each flip is possible. In fact,
$P(\text{5 heads in 5 flips})=\left(\frac12\right)^5=\frac{1}{32}.$

2. Conclude whether the observation proves bias

Since the probability is small but not zero, the result may happen by chance. Therefore it is not conclusive that the coin is biased.

1. Use the card from Bag A as the tens digit and the card from Bag B as the units digit

For each outcome, the number formed is:
$\text{number}=10\times(\text{Bag A digit})+(\text{Bag B digit}).$
For example, Bag A $=2$ and Bag B $=4$ gives $24$.

2. Complete every cell of the possibility diagram

Combine Bag A digits $1,2,3$ with Bag B digits $4,5,7$:

1. Count the favourable odd outcomes and divide by the total number of outcomes

From the possibility diagram, the possible numbers are
$14,24,34,15,25,35,17,27,37$.

Odd numbers end in $5$ or $7$:
$15,25,35,17,27,37$.

So there are $6$ odd numbers out of $9$ possible numbers.

$P(\text{odd})=\frac{6}{9}=\frac{2}{3}$

1. Count the favourable perfect-square outcome and divide by the total number of outcomes

From the possibility diagram, the possible numbers are
$14,24,34,15,25,35,17,27,37$.

The only perfect square is $25$.

Total number of possible outcomes $=9$.

$P(\text{perfect square})=\frac{1}{9}$

1. Identify the prime outcomes and form the probability

The possible numbers are
$14,24,34,15,25,35,17,27,37$.

The prime numbers are $17$ and $37$ only.

There are $9$ equally likely outcomes, so
$P(\text{prime number})=\frac{2}{9}.$

1. Identify the outcomes that are factors of 135

The possible numbers are
$14, 24, 34, 15, 25, 35, 17, 27, 37$.

Since
$135=1\times135=3\times45=5\times27=9\times15,$
the factors of $135$ among the possible numbers are
$15,\ 27,\ 45,\ 135,$
but only $15$ and $27$ are in the possibility diagram.

2. Form the probability and simplify the fraction

There are $9$ equally likely outcomes in total.

Number of favourable outcomes $=2$.

$P(\text{factor of }135)=\frac{2}{9}$