The probability that chris,izek and patience will pass examination are 2/3,5/8 and 3/4 respectively find that .

Question

(1) All the three will pass the examination.
(2) only one will pass the examination.
(3) only two will pass the examination.

Answer 1

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