Probability(not mutually exclusive events)
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The document contains examples of probability problems involving calculating the probability of events occurring. It provides the sample space, defines the events, calculates the probabilities of each event occurring individually and jointly to determine the probability of one event or the other occurring.
Probability(not mutually exclusive events)
- 2. Example-13: A die is rolled once, find the probability that face is an even number or a prime number. Solution: S = {1, 2, 3, 4, 5, 6} n (S) = 6 Let A = an even number A = {2, 4, 6} n (A) = 3 Let B = a prime number B = {2, 3, 5} n (B) = 3 Check thus events are not mutually exclusive therefore n Thus, ( ) ( ) ( ) 6 3 Sn An AP == ( ) ( ) ( ) 6 3 Sn Bn BP == 2BA = ( ) 1BA = ( ) ( ) ( ) 6 1 Sn BAn BAP = = ( ) ( ) ( ) ( ) 8333.0 6 5 6 1 6 3 6 3 ==−+=−+= BAPBPAPBAP
- 3. Example-14: A digit is selected at random from the first 12 natural numbers. Find the probability that the selected digit is multiple of 3 or multiple of 6. Solution: S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} n (S) = 12 Let A = digit is multiple of 3 A = {3, 6, 9, 12} n (A) = 4 Let B = digit is multiple of 6 B = {6, 12} n (B) = 2 Check thus events are not mutually exclusive. Thus, Therefore, ( ) ( ) ( ) 12 4 Sn An AP == ( ) ( ) ( ) 12 2 Sn Bn BP == 12,6BA = ( ) 2BAn = ( ) ( ) ( ) 12 2 Sn BAn BAP = = ( ) ( ) ( ) ( ) 3333.0 12 4 12 2 12 2 12 4 ==−+=−+= BAPBPAPBAP
- 4. Example-15: Two coins are tossed, find the probability that both faces are heads or at least one is head. Solution: S = {HH, HT, TH, TT} n (S) = 4 Let A = Both faces are heads A = {HH} n (A) = 1 Let B = At least one is head B = {HH, HT, TH} n (B) = 3 Check the events are not mutually exclusive. Thus, Therefore, ( ) ( ) ( ) 4 1 Sn An AP == ( ) ( ) ( ) 4 3 Sn Bn BP == HHBA = ( ) 1BAn = ( ) ( ) ( ) 4 1 Sn BAn BAP = = ( ) ( ) ( ) ( ) 75.0 4 3 4 1 4 3 4 1 ==−+=−+= BAPBPAPBAP
- 5. Example-16: A drum contains 8 bolts and 12 nuts. Half of the bolts and half of the nuts are rusted if one item is chosen at random, what is the probability that it is rusted or is a bolt? Solution: S = {B1 B2 B3 B4 B5 B6 B7 B8 N1 N2 N3 N4 N5 N6 N7 N8 N9 N10 N11 N12} n (S) = 20 Let A = it is rusted (half of the bolts and half of the nuts are rusted) A = {B1 B2 B3 B4 N1 N2 N3 N4 N5 N6} n (A) = 10 Let B = it is a bolt B = {B1 B2 B3 B4 B5 B6 B7 B8} n (B) = 8 Since, A B = {B1 B2 B3 B4} n (AB) = 4 Thus events are not mutually exclusive. Therefore, ( ) ( ) ( ) 20 10 Sn An AP == ( ) ( ) ( ) 20 8 Sn Bn BP == ( ) ( ) ( ) 20 4 Sn BAn BAP = = ( ) ( ) ( ) ( ) 7.0 20 14 20 4 20 8 20 10 ==−+=−+= BAPBPAPBAP
- 6. Example-17: A class contains 10 boys and 16 girls, half of the boys and half of the girls have black eyes. Find the probability that a student chosen at random is a girl or has black eyes. Solution: Girls Boys Total Black Eyes 8 5 13 Not Black Eyes 8 5 13 Total 16 10 26 n (S) = 26 Let A = Chosen is a girl n (A) = 16 Let B = Chosen has black eyes n (B) = 13 Since, Thus, events are not mutually exclusive. Therefore, ( ) ( ) ( ) 26 16 Sn An AP == ( ) ( ) ( ) 26 13 Sn 13n BP == ( ) 8BAn = ( ) ( ) ( ) 26 8 Sn BAn BAP = = ( ) ( ) ( ) ( ) 8077.0 26 21 26 8 26 13 26 16 BAPBPAPBAP ==−+=−+=