The probability that chris,izek and patience will pass examination are 2/3,5/8 and 3/4 respectively find that .
(1) All the three will pass the examination.
(2) only one will pass the examination.
(3) only two will pass the examination.
To find the probabilities for each scenario, we need to multiply the individual probabilities of passing for each person.
1) All three will pass the examination:
Probability = (2/3) * (5/8) * (3/4) = 30/96 = 5/16 or approximately 0.3125
2) Only one will pass the examination:
Chris passes while Izek and Patience fail: (2/3) * (3/8) * (1/4) = 6/96 = 1/16 or approximately 0.0625
Izek passes while Chris and Patience fail: (1/3) * (5/8) * (1/4) = 5/96 or approximately 0.0521
Patience passes while Chris and Izek fail: (1/3) * (3/8) * (3/4) = 9/96 = 3/32 or approximately 0.09375
Adding these probabilities together: 1/16 + 5/96 + 3/32 = 6/96 + 5/96 + 9/96 = 20/96 = 5/24 or approximately 0.2083
3) Only two will pass the examination:
Chris and Izek pass while Patience fails: (2/3) * (5/8) * (1/4) = 10/96 = 5/48 or approximately 0.1042
Chris and Patience pass while Izek fails: (2/3) * (3/8) * (3/4) = 18/96 = 3/16 or approximately 0.1875
Izek and Patience pass while Chris fails: (1/3) * (5/8) * (3/4) = 15/96 = 5/32 or approximately 0.15625
Adding these probabilities together: 5/48 + 3/16 + 5/32 = 10/96 + 18/96 + 15/96 = 43/96 or approximately 0.4479
Therefore, the probabilities are:
1) All three will pass the examination: 5/16 or approximately 0.3125
2) Only one will pass the examination: 5/24 or approximately 0.2083
3) Only two will pass the examination: 43/96 or approximately 0.4479