From the given data, the following Table is calculated:
| O | A | B | AB | Total | |
| RH+ | 39 | 31 | 8 | 9 | 87 |
| RH- | 5 | 10 | 9 | 6 | 30 |
| Total | 44 | 41 | 17 | 15 | 117 |
(a)
P(Not A) = 1 - P(A) = 1 - (41/117) = 1 - 0.3504 =
0.6496
(b)
P(A OR RH-) = P(A) + P(RH-) - P(A AND RH-)
= (41/117) + (30/117) - (10/117)
= 61/117 = 0.5214
(c)
P(RH+/O) = P(RH+ AND O)/P(O)
= 39/44 = 0.8864
(d)
P(RH+ AND O) = 39/117 = 0.3333
P(RH+) = 87/117 = 0.7436
P(O) = 44/117 = 0.3761
P(RH+) X P(O) = 0.2796
Since P(RH+ AND O) P(RH+) X P(O), we
conclude that the events"had Rh +" and "blood type O" are not
independent.
(e)
P(A OR B/RH+) = P(A OR B AND RH+)/P(RH+)
= 39/87 = 0.4483
(f)
P(Two subjects AB) = (15/117)2 = 0.0164
1. Use the data in the following table, which summarizes blood groups and Rh types for...
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