Question 22 of 22intermediate🔍 AnalyzeLong Answer4 marks

Which of the following arguments are correct and which are not correct? Give reasons for your answer.

(i) If two coins are tossed simultaneously there are three possible outcomes—two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is 1/31/3.

(ii) If a die is thrown, there are two possible outcomes—an odd number or an even number. Therefore, the probability of getting an odd number is 1/21/2.

(i)
Answer

The argument is not correct. When two coins are tossed simultaneously, the possible outcomes are HH, HT, TH, and TT. There are 4 possible outcomes, not 3. The outcome 'one of each' consists of two cases—HT and TH.

So,

P(two heads)=14P(\text{two heads}) = \frac{1}{4} P(two tails)=14P(\text{two tails}) = \frac{1}{4} P(one of each)=24=12P(\text{one of each}) = \frac{2}{4} = \frac{1}{2}

These three outcomes are not equally likely, so the probability is not 1/31/3 for each.

(ii)
Answer

The argument is correct. When a die is thrown, the possible outcomes are 1, 2, 3, 4, 5, 6, and each number has the same possibility of showing up (equally likely outcomes). The odd numbers are 1, 3, 5 (3 outcomes) and the even numbers are 2, 4, 6 (3 outcomes).

Therefore,

P(odd number)=Number of favourable outcomesTotal outcomes=36=12P(\text{odd number}) = \frac{\text{Number of favourable outcomes}}{\text{Total outcomes}} = \frac{3}{6} = \frac{1}{2}

Explanation

This question tests understanding of equally likely outcomes in probability. For part (i), students often incorrectly group HT and TH as a single outcome 'one of each', but these are two distinct outcomes. The probability formula P(E)=favourable outcomes/total outcomesP(E) = \text{favourable outcomes} / \text{total outcomes} requires all outcomes to be equally likely.

In part (ii), the classification into odd and even correctly divides the 6 equally likely outcomes into two groups of 3 each, making the probability calculation valid.

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