Example In a box there are three balls: red, green and blue. We - PowerPoint PPT Presentation

Example In a box there are three balls: red, green and blue. We draw one ball at random, replace it and draw a second ball. Describe the sample space for this experiment. How does the sample space change if we do not replace the first ball before drawing the second one?

Solution Define the following: R = draw a red ball G = draw a green ball B = draw a blue ball Ω = { ( R,R), (R,G), (R,B), (G,G), (G,R), (G,B), (B,B), (B,R), (B,G) } (with replacement) Ω = { ( R,G), (R,B), (G,R), (G,B), (B,R), (B,G) } (without replacement)

Example We toss the same coin three times. Calculate probabilities of the following events: A = „we get exactly two heads ” B = „we do not get two tails in a row ” C = „we get at least one tails ” D = „we get heads on the second throw ”

Solution Possible results of throwing a coin three times: TTT HHH THT HHT TTH HTH THH HTT

Solution TTT HHH A = „we get exactly two heads ” THT HHT TTH HTH THH HTT P(A) = 3/8

Solution TTT HHH THT HHT TTH HTH B = „we do not get two tails in a row ” THH HTT P(B) = 5/8

Solution TTT HHH THT HHT C = „we get at least one tails ” TTH HTH THH HTT P(C) = 7/8

Solution TTT HHH THT HHT D = „we get heads on the second throw ” TTH HTH THH HTT P(D) = 4/8 = 1/2

Example • In a certain region weather on any day can be classified as „ good ”, „ moderate ” and „ bad ”. Probability of each type of weather occuring on any day is 0.2, 0.5 and 0.3 respectively. • If the weather is good, probability of rain is 0.3. If the weather is moderate, probability of rain is 0.5. If the weather is bad probability of rain is 0.9. • What is the probability that it will rain on any day?

Solution 𝑄 𝑕𝑝𝑝𝑒 𝑥𝑓𝑏𝑢ℎ𝑓𝑠 = 𝑄 𝐶 1 = 0.2 𝑄 𝑛𝑝𝑒𝑓𝑠𝑏𝑢𝑓 𝑥𝑓𝑏𝑢ℎ𝑓𝑠 = 𝑄 𝐶 2 = 0.5 𝑄 𝑐𝑏𝑒 𝑥𝑓𝑏𝑢ℎ𝑓𝑠 = 𝑄 𝐶 3 = 0.3 𝑄 𝑠𝑏𝑗𝑜 𝑕𝑗𝑤𝑓𝑜 𝑕𝑝𝑝𝑒 𝑥𝑓𝑏𝑢ℎ𝑓𝑠 = 𝑄 𝐵|𝐶 1 = 0.3 𝑄 𝑠𝑏𝑗𝑜 𝑕𝑗𝑤𝑓𝑜 𝑛𝑝𝑒𝑓𝑠𝑏𝑢𝑓 𝑥𝑓𝑏𝑢ℎ𝑓𝑠 = 𝑄 𝐵|𝐶 2 = 0.5 𝑄 𝑠𝑏𝑗𝑜 𝑕𝑗𝑤𝑓𝑜 𝑐𝑏𝑒 𝑥𝑓𝑏𝑢ℎ𝑓𝑠 = 𝑄 𝐵|𝐶 3 = 0.9 𝑄 𝑠𝑏𝑗𝑜 = 𝑄 𝐵 = 𝑄 𝐵|𝐶 1 𝑄 𝐶 1 + 𝑄 𝐵|𝐶 2 𝑄 𝐶 2 + 𝑄 𝐵|𝐶 3 𝑄(𝐶 3 ) = = 0.3 × 0.2 + 0.5 × 0.5 + 0.9 × 0.3 = 0.58

Solution 0.2 0.5 0.3 good bad moderate 0.9 0.1 0.7 0.5 0.5 0.3 rain dry rain rain dry dry

Solution 0.2 0.5 0.3 good bad moderate 0.9 0.1 0.7 0.5 0.5 0.3 rain dry rain rain dry dry P(rain) = 0.3 × 0.2 + 0.5 × 0.5 + 0.9 × 0.3 = 0.58

Example In a company there are three categories of employees: admin (30% of employees), specialist (55% of employees) and executive (15% of employees). Probability of an employee receiving a bonus is equal to 0.2 for admin, 0.4 for specialist and 0.6 for executive. An employee was chosen at random. If we know that this employee has received a bonus, what is the probability that they belong to each of the three categories?

Solution 𝑄 𝑏𝑒𝑛𝑗𝑜 = 𝑄 𝐶 1 = 0.3 𝑄 𝑡𝑞𝑓𝑑𝑗𝑏𝑚𝑗𝑡𝑢 = 𝑄 𝐶 2 = 0.55 𝑄 𝑓𝑦𝑓𝑑𝑣𝑢𝑗𝑤𝑓 = 𝑄 𝐶 3 = 0.15 𝑄 𝑐𝑝𝑜𝑣𝑡 𝑕𝑗𝑤𝑓𝑜 𝑏𝑒𝑛𝑗𝑜 = 𝑄 𝐵|𝐶 1 = 0.2 𝑄 𝑐𝑝𝑜𝑣𝑡 𝑕𝑗𝑤𝑓𝑜 𝑡𝑞𝑓𝑑𝑗𝑏𝑚𝑗𝑡𝑢 = 𝑄 𝐵|𝐶 2 = 0.4 𝑄 𝑐𝑝𝑜𝑣𝑡 𝑕𝑗𝑤𝑓𝑜 𝑓𝑦𝑓𝑑𝑣𝑢𝑗𝑤𝑓 = 𝑄 𝐵|𝐶 3 = 0.6 𝑄 𝐵 = 0.3 × 0.2 + 0.55 × 0.4 + 0.15 × 0.6 = 0.37

Solution 𝑄 𝐶 1 |𝐵 = 𝑄(𝐵|𝐶 1 ) × 𝑄(𝐶 1 ) = 0.2 × 0.3 = 0.162 𝑄(𝐵) 0.37 𝑄 𝐶 2 |𝐵 = 𝑄(𝐵|𝐶 2 ) × 𝑄(𝐶 2 ) = 0.4 × 0.55 = 0.595 𝑄(𝐵) 0.37 𝑄 𝐶 3 |𝐵 = 𝑄(𝐵|𝐶 3 ) × 𝑄(𝐶 3 ) = 0.6 × 0.15 = 0.243 𝑄(𝐵) 0.37