Question

Difficulty: MediumExperimental and Theoretical Probability

A box contains 3030 tickets numbered 11 to 3030. In a probability experiment, a ticket is drawn at random from the box and its number recorded before being replaced. This trial is performed 150150 times, and a ticket with a number that is a multiple of 44 is recorded 4545 times. What is the absolute difference between the experimental probability and the theoretical probability of selecting a ticket bearing a multiple of 44?

  1. 115\frac{1}{15}Answer
  2. B
    310\frac{3}{10}
  3. C
    730\frac{7}{30}
  4. D
    815\frac{8}{15}

Answer

The absolute difference between the experimental probability and the theoretical probability is 115\frac{1}{15}.
The theoretical probability of drawing a multiple of 4 is 730\frac{7}{30} since there are 7 favorable tickets (4,8,12,16,20,24,284, 8, 12, 16, 20, 24, 28) out of 30 total tickets. The experimental probability from 150 draws is 45150=930\frac{45}{150} = \frac{9}{30}. Taking the absolute difference gives 930730=230=115\frac{9}{30} - \frac{7}{30} = \frac{2}{30} = \frac{1}{15}.

Step-by-Step Solution

1
Determine the theoretical probability
Theoretical probability P(T)=730P(T) = \frac{7}{30}
The multiples of 44 between 11 and 3030 are 4,8,12,16,20,24,284, 8, 12, 16, 20, 24, 28, giving 77 favorable outcomes out of 3030 possible outcomes.
2
Determine the experimental probability
Experimental probability P(E)=45150=310P(E) = \frac{45}{150} = \frac{3}{10}
The event occurred 4545 times out of 150150 experimental trials.
3
Calculate the absolute difference between P(E)P(E) and P(T)P(T)
P(E)P(T)=930730=230=115|P(E) - P(T)| = |\frac{9}{30} - \frac{7}{30}| = \frac{2}{30} = \frac{1}{15}
Express both fractions with a common denominator of 3030 and subtract the smaller probability from the larger.

Key Concept

Experimental probability is calculated from trial data (favorable trials divided by total trials), whereas theoretical probability is calculated from expected sample space outcomes (favorable outcomes divided by total sample space size).
Estimated Time:1m 30s
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