Probability of an Event • Introduction ○ We first study Pierre-Simon Laplace’s classical theory of probability ○ which he introduced in the 18th century, when he analyzed games of chance. • Experiment ○ A procedure that yields one of a given set of possible outcomes. • Sample space ○ The sample space of the experiment is the set of possible outcomes. • Event ○ An event is a subset of the sample space. • Probability ○ If S is a finite sample space of equally likely outcomes and E is an event ○ Then the probability of E is p(E)=|E|/|S| ○ For every event E, we have 0≤p(E)≤1 ○ This follows directly from the definition because ○ 0≤p(E)=|E|/|S| ≤|S|/|S| ≤1, since 0≤|E|≤|S| Applying Laplace’s Definition • Example 1 ○ An urn contains four blue balls and five red balls. ○ What is the probability that a ball chosen from the urn is blue? ○ The probability that the ball is chosen is 4/9 ○ since there are nine possible outcomes, and four of these produce a blue ball. • Example 2 ○ What is the probability that when two dice are rolled, the sum of the numbers on the two dice is 7? ○ By the product rule there are 62 = 36 possible outcomes. ○ Six of these sum to 7. ○ Hence, the probability of obtaining a 7 is 6/36 = 1/6. • Example 3 ○ In a lottery, a player wins a large prize when they pick four digits that match, in correct order, four digits selected by a random mechanical process. ○ What is the probability that a player wins the prize? ○ By the product rule there are 104 = 10,000 ways to pick four digits. ○ Since there is only 1 way to pick the correct digits, ○ the probability of winning the large prize is 1/10,000 = 0.0001. • Example 4 ○ A smaller prize is won if only three digits are matched. ○ What is the probability that a player wins the small prize? ○ If exactly three digits are matched, one of the four digits must be incorrect and the other three digits must be correct. ○ For the digit that is incorrect, there are 9 possible choices. ○ Hence, by the sum rule, there a total of 36 possible ways to choose four digits that match exactly three of the winning four digits. ○ The probability of winning the small price is § 36/10,000 = 9/2500 = 0.0036 • Example 5 ○ There are many lotteries that award prizes to people who correctly choose a set of six numbers out of the first n positive integers, where n is usually between 30 and 60. ○ What is the probability that a person picks the correct six numbers out of 40? ○ The number of ways to choose six numbers out of 40 is § C(40,6) = 40!/(34!6!) = 3,838,380. ○ Hence, the probability of picking a winning combination is § 1/ 3,838,380 ≈ 0.00000026. • Example 6 ○ What is the probability that the numbers 11, 4, 17, 39, and 23 are drawn in that order from a bin with 50 balls labeled with the numbers 1,2, …, 50 if ○ The ball selected is not returned to the bin. § Sampling without replacement: § The probability is 1/254,251,200 since there are § 50 ∙49 ∙47 ∙46 ∙45 = 254,251,200 ways to choose the five balls. ○ The ball selected is returned to the bin before the next ball is selected. § Sampling with replacement: § The probability is 1/〖50〗^5 = 1/312,500,000 since 〖50〗^5 = 312,500,000. The Probability of Complements and Unions of Events • Theorem 1 ○ Let E be an event in sample space S. ○ The probability of the complementary event of E:E ̅ = S − E is given by ○ p(E ̅ )=1−p(E) • Proof ○ Using the fact that |E ̅ |=|S|−|E| ○ p(E ̅ )=(|S|−|E|)/|S| =1−|E|/|S| =1−p(E) • Example ○ A sequence of 10 bits is chosen randomly. ○ What is the probability that at least one of these bits is 0? ○ Let E be the event that at least one of the 10 bits is 0. ○ Then E ̅ is the event that all of the bits are 1s. ○ The size of the sample space S is 210. Hence, ○ p(E)=1−p(E ̅ )=1−|E ̅ |/|S| =1−1/2^10 =1−1/1024=1023/1024 • Theorem 2 ○ Let E_1 and E_2 be events in the sample space S. Then ○ p(E_1∪E_2 )=p(E_1 )+p(E_2 )−p(E_1∩E_2 ) • Proof ○ Given the inclusion-exclusion formula from Section 2.2 ○ |A ∪ B| = |A| + |B| − |A ∩ B|, it follows that § p(E_1∪E_2 )=|E_1∪E_2 |/|S| § =(|E_1 |+|E_2 |−|E_1∩E_2 |)/|S| § =|E_1 |/|S| +|E_2 |/|S| −|E_1∩E_2 |/|S| § =p(E_1 )+p(E_2 )−p(E_1∩E_2 ) • Example ○ What is the probability that a positive integer selected at random from the set of positive integers not exceeding 100 is divisible by either 2 or 5? ○ Let E_1 be the event that the integer is divisible by 2 ○ Let E_2 be the event that it is divisible 5. ○ Then the event that the integer is divisible by 2 or 5 is E_1∪E_2 ○ And E_1∩E_2 is the event that it is divisible by 2 and 5. ○ It follows that: ○ p(E_1∪E_2 )=p(E_1 )+p(E_2 )−p(E_1∩E_2 )=50/100+20/100−10/100=3/5 Monty Hall Puzzle • You are asked to select one of the three doors to open. • There is a large prize behind one of the doors and if you select that door, you win the prize. • After you select a door, the game show host opens one of the other doors (which he knows is not the winning door). • The prize is not behind the door and he gives you the opportunity to switch your selection. • Should you switch? • You should switch. • The probability that your initial pick is correct is 1/3. • This is the same whether or not you switch doors. • But since the game show host always opens a door that does not have the prize, • If you switch the probability of winning will be 2/3 • Because you win if your initial pick was not the correct door • And the probability your initial pick was wrong is 2/3.
