Da experiment invloved tossing three coins simultaneously da experient was carried out 100 times, it was noted dat three heads occurred 40 times. Wut is da difference between da experimental prbablity of getting three heads n its theroretical probability rite da answer in simplest form of fraction

1 answer

To find the experimental probability of getting three heads and the theoretical probability, let's calculate both.

  1. Experimental Probability:

    • The experiment was carried out 100 times, and three heads occurred 40 times.
    • Therefore, the experimental probability \( P(E) \) of getting three heads is: \[ P(E) = \frac{\text{Number of times three heads occurred}}{\text{Total number of tosses}} = \frac{40}{100} = \frac{2}{5} \]
  2. Theoretical Probability:

    • When tossing three coins, the possible outcomes are:
      • HHH, HHT, HTH, HHT, THH, HTT, THT, TTH, TTT
    • There are a total of \( 2^3 = 8 \) possible outcomes. Only one of these outcomes is three heads (HHH).
    • Therefore, the theoretical probability \( P(T) \) of getting three heads is: \[ P(T) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{8} \]
  3. Finding the Difference:

    • Now, we need to find the difference between the experimental probability and the theoretical probability: \[ \text{Difference} = P(E) - P(T) = \frac{2}{5} - \frac{1}{8} \]
    • To subtract these fractions, find a common denominator. The least common multiple of 5 and 8 is 40.
    • Convert each fraction: \[ \frac{2}{5} = \frac{2 \times 8}{5 \times 8} = \frac{16}{40} \] \[ \frac{1}{8} = \frac{1 \times 5}{8 \times 5} = \frac{5}{40} \]
    • Now subtract the fractions: \[ \text{Difference} = \frac{16}{40} - \frac{5}{40} = \frac{16 - 5}{40} = \frac{11}{40} \]

Thus, the difference between the experimental probability of getting three heads and its theoretical probability is \( \frac{11}{40} \).