The probability that a candidate chosen at random from among these two sets of candidates will have selected questions 1, 2 and 3 is 0.4179.
The probability that a candidate chosen at random from among these two sets of candidates will have selected questions 1, 2 and 3 will ba calculated thus:
= [(9C6 /12C9) × 1/2] + [(9C7)/12C10) × 1/2]
= 42/220 + 18/66
= 0.1909 + 0.2727
= 0.4179
In conclusion, the probability is 0.4179.
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