Question 09.
A student says that, if you throw a die, it will show up 1 or not 1. Therefore, the probability of getting 1 and the probability of getting not 1 each is equal to $\frac{1}{2}$ . Is this correct ? Give reasons.

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Solution

No, this is not correct.
Suppose we throw a die, then total number of outcomes = 6
Possible outcomes = 1 or 2 or 3 or 4 or 5 or 6
$∴$ Probability of getting $1 = \frac{1}{6}$
Now, Probability of getting not 1 = 1 - Probability of getting 1
$= 1 - \frac{1}{6} = \frac{5}{6}$

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The probability of getting number 6 in a throw of a die is $\frac{1}{6} .$ Similarly, the probability of getting a number 5 is $\frac{1}{5} .$

Q. Question 11.
If you toss a coin 6 times and it comes down heads on each occasion. Can you say that the probability of getting a head is 1 ? Given reasons.

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.

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\frac{1}{2}

Q. In a single throw of a die, the probability of getting a number greater than or equal to 1 is __.

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Equally Likely Events

Standard IX Mathematics

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Question 09. A student says that, if you throw a die, it will show up 1 or not 1. Therefore, the probability of getting 1 and the probability of getting not 1 each is equal to 12. Is this correct ? Give reasons.

No, this is not correct. Suppose we throw a die, then total number of outcomes = 6 Possible outcomes = 1 or 2 or 3 or 4 or 5 or 6 ∴ Probability of getting 1=16 Now, Probability of getting not 1 = 1 - Probability of getting 1 =1−16=56

Question 09.
A student says that, if you throw a die, it will show up 1 or not 1. Therefore, the probability of getting 1 and the probability of getting not 1 each is equal to $\frac{1}{2}$ . Is this correct ? Give reasons.

No, this is not correct.
Suppose we throw a die, then total number of outcomes = 6
Possible outcomes = 1 or 2 or 3 or 4 or 5 or 6
$∴$ Probability of getting $1 = \frac{1}{6}$
Now, Probability of getting not 1 = 1 - Probability of getting 1
$= 1 - \frac{1}{6} = \frac{5}{6}$