Skip to main content

Three students solving a problem

Short answer

P(A)=1/2, P(B)=3/4, P(C)=1/4. (a) At least one solves: 1 − (1/2×1/4×3/4)=29/32=0.906. (b) None solves: (1/2×1/4×3/4)=3/32=0.094. (c) Exactly two: (AB not C) + (AC not B) + (BC not A) = 9/32+1/32+3/32=13/32≈0.406.

A problem in statistics is given to three students A, B and C whose chances of solving are 1/2, 3/4 and 1/4 respectively. If all of them try independently, what is the probability that a) at least one of them will solve it b) none of them can solve it c) exactly two of them can solve it

Let A, B, C be the events that the student solves the problem.

P(A) = 1/2, P(B) = 3/4, P(C) = 1/4, so P(A′) = 1/2, P(B′) = 1/4, P(C′) = 3/4. The students try independently, so probabilities multiply.

a) At least one solves it

P = 1 − P(none solves)
  = 1 − (1/2)(1/4)(3/4)
  = 1 − 3/32
  = 29/32 = 0.906

b) None can solve it

P = P(A′) P(B′) P(C′) = (1/2)(1/4)(3/4) = 3/32 = 0.094

c) Exactly two solve it

Three cases: AB solve and C does not, AC solve and B does not, BC solve and A does not.

A, B, not C: (1/2)(3/4)(3/4) = 9/32
A, C, not B: (1/2)(1/4)(1/4) = 1/32
B, C, not A: (1/2)(3/4)(1/4) = 3/32
Total = 13/32 = 0.406

Answer: (a) 29/32 = 0.906, (b) 3/32 = 0.094, (c) 13/32 = 0.406.

✓ Answer checked by Claude Sonnet 5.5

Related questions

All Probability and Statistics chapter 1 questions →