Q25(i):
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 $\frac{1}{3}$.

Solution :

Given: Two coins are tossed simultaneously. The argument states that there are three possible outcomes: two heads (HH), two tails (TT), or one of each (HT or TH). Based on this, it claims the probability of each outcome is $\frac{1}{3}$.

To Find: Determine if the argument is correct or incorrect and provide a mathematical justification.

Step 1: Identifying the Sample Space
When two fair coins are tossed simultaneously, each coin has two possible outcomes: Head ($H$) or Tail ($T$). Let the outcomes of the first coin be $C_1$ and the second coin be $C_2$. The total number of possible outcomes is determined by the Fundamental Counting Principle: $2 \times 2 = 4$.
The sample space $S$ is the set of all possible elementary events:
$S = \{HH, HT, TH, TT\}$
Where:
$HH$ = Head on both coins
$HT$ = Head on the first coin, Tail on the second
$TH$ = Tail on the first coin, Head on the second
$TT$ = Tail on both coins

Step 2: Analyzing the Argument's Outcomes
The argument suggests three outcomes: "two heads", "two tails", or "one of each".
Let us map these to our sample space $S$:
- Two heads: $\{HH\}$ (1 outcome)
- Two tails: $\{TT\}$ (1 outcome)
- One of each: $\{HT, TH\}$ (2 outcomes)

Step 3: Calculating Probabilities
Using the classical definition of probability, $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$.
The total number of outcomes in the sample space is $n(S) = 4$.

Calculating the probability for each category defined in the argument:

1. Probability of two heads ($P_{HH}$):
$P(HH) = \frac{1}{4}$

2. Probability of two tails ($P_{TT}$):
$P(TT) = \frac{1}{4}$

3. Probability of one of each ($P_{one\_of\_each}$):
$P(HT \text{ or } TH) = \frac{2}{4} = \frac{1}{2}$

Step 4: Comparison and Conclusion
The argument claims that the probability for each of the three outcomes is $\frac{1}{3}$. However, our calculations show:
$P(HH) = \frac{1}{4} \neq \frac{1}{3}$
$P(TT) = \frac{1}{4} \neq \frac{1}{3}$
$P(one\_of\_each) = \frac{1}{2} \neq \frac{1}{3}$

The error in the argument lies in the assumption that the three outcomes are "equally likely". In probability theory, the classical formula $P(E) = \frac{1}{n}$ only applies when all elementary events in the sample space are equally likely. Since the event "one of each" consists of two distinct elementary outcomes ($HT$ and $TH$), it is twice as likely to occur as "two heads" or "two tails".

Final Answer: The argument is incorrect. The outcomes are not equally likely; the probability of getting two heads is $\frac{1}{4}$, two tails is $\frac{1}{4}$, and one of each is $\frac{1}{2}$.

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Q25(i): 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 $\frac{1}{3}$.

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CBSE Solutions for Class 10 Mathematics Probability

Chapters in CBSE - Class 10 Mathematics

Other Subjects in CBSE - Class 10

Q25(i):
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 $\frac{1}{3}$.