树状图练习题 - 掌握连续事件概率计算和树状图应用
以下是3道精选练习题，涵盖树状图的各种应用场景，包括独立事件、条件概率和放回与不放回情况。
Kaan takes part in two cycle races. The probability that he wins the first race is 0.6. The probability that he wins the second race is 0.7. Work out the probability that Kaan wins at least one race.
解答过程：
Probability of winning at least one race = \( 1 - P(\text{loses both}) \)
\( P(\text{loses first}) = 1 - 0.6 = 0.4 \)
\( P(\text{loses second}) = 1 - 0.7 = 0.3 \)
\( P(\text{loses both}) = 0.4 \times 0.3 = 0.12 \)
\( P(\text{wins at least one}) = 1 - 0.12 = 0.88 \)
Aleena has two coins. One is fair, with a head on one side and a tail on the other. The second is a trick coin and has a tail on both sides. Aleena picks up one of the coins at random and flips it.
a) Find the probability that it lands heads up.
b) Given that it lands tails up, find the probability that she picked up the fair coin.
解答过程：
a) \( P(\text{heads}) = P(\text{fair coin}) \times P(\text{heads on fair}) + P(\text{trick coin}) \times P(\text{heads on trick}) = 0.5 \times 0.5 + 0.5 \times 0 = 0.25 \)
b) \( P(\text{tails}) = 0.5 \times 0.5 + 0.5 \times 1 = 0.75 \)
\( P(\text{fair coin} | \text{tails}) = \frac{0.5 \times 0.5}{0.75} = \frac{1}{3} \)
A box of jelly beans contains 7 sweet flavours and 3 sour flavours. Two of the jelly beans are taken one after the other and eaten. Emilia wants to find the probability that both jelly beans eaten are sweet, given that at least one of them is. Identify Emilia's mistake and find the correct probability.
解答过程：
Emilia's mistake：She assumed the draws are with replacement (used \( \frac{7}{10} \times \frac{7}{10} \)), but they are without replacement.
Correct steps：
\( P(\text{both sweet}) = \frac{7}{10} \times \frac{6}{9} = \frac{7}{15} \)
\( P(\text{at least one sweet}) = 1 - P(\text{both sour}) = 1 - \frac{3}{10} \times \frac{2}{9} = \frac{14}{15} \)
\( P(\text{both sweet} | \text{at least one sweet}) = \frac{\frac{7}{15}}{\frac{14}{15}} = \frac{1}{2} \)
一个袋子里有5个红球和3个蓝球。从袋子里不放回地抽取两个球。求：
a) 两个球都是红球的概率
b) 第一个球是红球且第二个球是蓝球的概率
c) 至少有一个球是红球的概率
解答过程：
a) 两红球：\( \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14} \)
b) 红后蓝：\( \frac{5}{8} \times \frac{3}{7} = \frac{15}{56} \)
c) 至少一红：1 - 两蓝概率 = \( 1 - \frac{3}{8} \times \frac{2}{7} = 1 - \frac{6}{56} = \frac{50}{56} = \frac{25}{28} \)
掷两个骰子，已知至少有一个骰子显示6，求两个骰子点数之和为8的概率。
解答过程：
至少一个6的情况（11种）：
(1,6), (2,6), (3,6), (4,6), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)
和为8的情况（2种）：
(2,6), (6,2)
\( P(\text{sum}=8 | \text{at least one 6}) = \frac{2}{11} \)