MHT CET202522 Apr 2025Evening ShiftMathematicsProbabilityActual
Four defective oranges are accidentally mixed with sixteen good ones. Three oranges are drawn from the mixed lot. The probability distribution of defective oranges is
Options
- A( array |c|c|c|c|c| X & 0 & 1 & 2 & 3 P ( X =x) & 28 57 & 8 95 & 8 19 & 1 285 array )
- B( array |c|c|c|c|c| X & 0 & 1 & 2 & 3 P ( X =x) & 28 57 & 8 19 & 8 95 & 1 285 array )
- C( array |c|c|c|c|c| X & 0 & 1 & 2 & 3 P ( X =x) & 28 57 & 8 95 & 1 285 & 8 19 array )
- D( array |c|c|c|c|c| X & 0 & 1 & 2 & 3 P ( X =x) & 1 285 & 8 95 & 8 19 & 28 57 array )
Correct answer
B. ( array |c|c|c|c|c| X & 0 & 1 & 2 & 3 P ( X =x) & 28 57 & 8 19 & 8 95 & 1 285 array )
Step-by-step solution
The lot contains 4 defective and 16 good oranges, totaling 20 oranges. Let X represent the number of defective oranges when drawing 3 without replacement. The total number of ways to draw 3 oranges is 20 3 = 1140 . For X = 0 : Probability is 4 0 16 3 1140 = 1 560 1140 = 28 57 . For X = 1 : Probability is 4 1 16 2 1140 = 4 120 1140 = 8 19 . For X = 2 : Probability is 4 2 16 1 1140 = 6 16 1140 = 8 95 . For X = 3 : Probability is 4 3 16 0 1140 = 4 1 1140 = 1 285 . The probability distribution matches Option B .