Let's go through each question one by one:
Question 1
Theoretical vs. Experimental Probability:
- Experimental Probability: This is based on the actual results of an experiment. In this case, Pashmina flipped the coin 20 times and got heads 15 times, which gives an experimental probability of \( P(\text{heads}) = \frac{15}{20} = \frac{3}{4} \).
- Theoretical Probability: This assumes an ideal situation where there is an equal chance for each outcome. For a fair double-sided coin, the theoretical probability of landing on heads is \( P(\text{heads}) = \frac{1}{2} \).
Answer: Theoretical probability that it will land on heads:
- Correct Response: \( \frac{1}{2} \)
Question 2
Talia's Spinner Probability:
- The letters in “Mississippi” are: M, I, S, S, I, S, S, I, P, P, I.
- The unique letters are M, I, S, and P. Letter 'I' appears 4 times, 'S' appears 4 times, 'P' appears 2 times.
- Total occurrences: 11 letters.
- The probability of landing on any letter in "Mississippi" (which includes I, S, and P) is based on the number of times those letters appear.
Letters in Mississippi: 4 I's + 4 S's + 2 P's = 10 occurrences of the letters that count.
Total letters on the spinner = 26.
Theoretical probability: \(P(\text{A letter in "Mississippi"}) = \frac{10}{26} = \frac{5}{13} \).
To simplify:
- 10 and 26 can be simplified to the expression \( \frac{5}{13} \).
Thus, if we take the response options into account,
Correct Response: \( \frac{2}{13} \)
Question 3
Comparing Experimental vs. Theoretical Probability:
- Sadiq's Experimental Probability: He rolled an even number 7 times out of 10, hence \( P(\text{even}) = \frac{7}{10} \).
- Theoretical Probability: The even numbers on a six-sided die are 2, 4, and 6. Therefore, the theoretical probability \( P(\text{even}) = \frac{3}{6} = \frac{1}{2} \).
Since \( \frac{7}{10} > \frac{1}{2} \), the experimental probability is greater.
Correct Response: The experimental probability was greater than the theoretical probability.
Question 4
Probability of Rolling Odd Numbers: A six-sided die has 3 odd numbers: 1, 3, and 5. The probability of rolling an odd number in a single roll is \( \frac{3}{6} = \frac{1}{2} \).
For three independent rolls, the probability of rolling odd three times in a row is calculated as: \[ P(\text{odd})^{3} = \left(\frac{1}{2}\right)^3 = \frac{1}{8} \].
Correct Response: \( \frac{1}{8} \)
Question 5
Probability of Rolling Two Number Cubes and Getting a Sum of 6 or 11:
- To find the valid combinations:
- Sum of 6: (1,5), (2,4), (3,3), (4,2), (5,1) – 5 combinations.
- Sum of 11: (5,6), (6,5) – 2 combinations.
The total successful outcomes = 5 (for sum of 6) + 2 (for sum of 11) = 7. Total possible outcomes when rolling two dice = \( 6 \times 6 = 36 \).
Thus, \[ P(\text{sum of 6 or 11}) = \frac{7}{36} \]
Correct Response: None of the provided responses is exactly correct; however, it should have been derived through \( \frac{5}{36} + \frac{2}{36} \) for our specific case.
Best Match Based on the Response Options:
\( \frac{6}{36} + \frac{2}{36} \) which equals \( \frac{7}{36} \) (if it were to encapsulate it), thus the best prediction solely bases on exploratory responses should also be \( \frac{5}{36} + \frac{2}{36} \).
Summary of Answers
- \( \frac{1}{2} \)
- \( \frac{2}{13} \)
- The experimental probability was greater than the theoretical probability.
- \( \frac{1}{8} \)
- \( \frac{5}{36} + \frac{2}{36} \) as the correct calculation for sum - or any indicating \( \frac{7}{36} \) as best understood in variants.
These responses summarise both calculating and understanding the outcomes based on contextual probability analysis.