If the probability of obtaining H was 0.2, find: i) The total number of times the die was thrown. ii) The probability of obtaining at least H

Here's the solution to the probability problem.

First, let's list the given frequencies from the table:

• Frequency of 1: 33
• Frequency of 2: 40
• Frequency of 3: 49
• Frequency of 4: $x$
• Frequency of 5: 42
• Frequency of 6: 37

The problem states: "If the probability of obtaining H was 0.2". Given the context of a die roll, 'H' is likely a typographical error for one of the numerical outcomes. We need to find which outcome, when its probability is set to 0.2, results in integer values for the unknown frequency $x$ and the total number of throws.

Let $N$ be the total number of times the die was thrown.
The sum of all frequencies is $N = 33 + 40 + 49 + x + 42 + 37 = 201 + x$.

We test the known frequencies:

• If $P(1) = 0.2 \implies \frac{33}{N} = 0.2 \implies N = 165$. Then $x = 165 - 201 = -36$ (invalid).
• If $P(2) = 0.2 \implies \frac{40}{N} = 0.2 \implies N = 200$. Then $x = 200 - 201 = -1$ (invalid).
• If $P(3) = 0.2 \implies \frac{49}{N} = 0.2 \implies N = 245$. Then $x = 245 - 201 = 44$ (valid).
• If $P(5) = 0.2 \implies \frac{42}{N} = 0.2 \implies N = 210$. Then $x = 210 - 201 = 9$ (valid).
• If $P(6) = 0.2 \implies \frac{37}{N} = 0.2 \implies N = 185$. Then $x = 185 - 201 = -16$ (invalid).

Both $P(3)=0.2$ and $P(5)=0.2$ lead to valid integer frequencies. Given the common occurrence of misinterpretations in handwritten text, we will proceed with the assumption that 'H' refers to the outcome '5', as it is a plausible misreading.

Assumption: The letter 'H' in the question "If the probability of obtaining H was 0.2" is a typographical error and refers to the outcome '5' from the die roll.