约翰抛了五次硬币，连续五次正面朝上。这是否能断定这枚硬币是有偏的（不均匀的）？请清晰地解释你的答案。 从袋子 A 中随机抽取一张卡片，然后再从袋子 B 中随机抽取另一张卡片，组成一个两位数。袋子 A 包含三张标有 '1'、'2' 和 '3' 的卡片。袋子 B 包含三张标有 '4'、'5' 和 '7' 的卡片。例如，分别从袋子 A 和袋子 B 中抽取 '1' 和 '5' 将组成数字 '15'。 以最简分数形式，求组成的两位数是以下情况的概率： 完成下面的可能性图。 奇数， 答案 ... 完全平方数， 答案 ... 质数， 答案 ... 135 的因数。 答案 ...

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 $=2$.

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