CBSE Class 9 Mathematics Chapter 15 Probability NCERT Solutions

The total number of balls played by the batswoman is given as 30.

The number of times she hit a boundary is 6.

To find the number of balls in which she did not hit a boundary, we subtract the number of boundaries hit from the total number of balls played:

Number of balls without a boundary = Total balls - Number of boundaries

$$30\; -\; 6\; =\; 24$$

The probability of an event is calculated as the ratio of the number of favorable outcomes to the total number of possible outcomes.

In this case, the favorable outcome is the event that the batswoman did not hit a boundary.

P(she did not hit a boundary) = $$\backslash frac\{\backslash text\{Number\; of\; balls\; without\; a\; boundary\}\}\{\backslash text\{Total\; number\; of\; balls\; played\}\}$$

$$P(\backslash text\{not\; hit\; boundary\})\; =\; \backslash frac\{24\}\{30\}$$

Simplifying the fraction:

$$\backslash frac\{24\}\{30\}\; =\; \backslash frac\{4\; \backslash times\; 6\}\{5\; \backslash times\; 6\}\; =\; \backslash frac\{4\}\{5\}$$

Therefore, the probability that the batswoman did not hit a boundary is $\backslash frac\{4\}\{5\}$.

The total number of families selected is 1500.

We need to compute the probability for each case:

1. Probability of a family having 2 girls:

   Number of families with 2 girls = 475

   P(2 girls) = $$\backslash frac\{\backslash text\{Number\; of\; families\; with\; 2\; girls\}\}\{\backslash text\{Total\; number\; of\; families\}\}\; =\; \backslash frac\{475\}\{1500\}$$

   Simplifying the fraction:

   $$\backslash frac\{475\}\{1500\}\; =\; \backslash frac\{19\; \backslash times\; 25\}\{60\; \backslash times\; 25\}\; =\; \backslash frac\{19\}\{60\}$$

2. Probability of a family having 1 girl:

   Number of families with 1 girl = 814

   P(1 girl) = $$\backslash frac\{\backslash text\{Number\; of\; families\; with\; 1\; girl\}\}\{\backslash text\{Total\; number\; of\; families\}\}\; =\; \backslash frac\{814\}\{1500\}$$

   Simplifying the fraction:

   $$\backslash frac\{814\}\{1500\}\; =\; \backslash frac\{407\; \backslash times\; 2\}\{750\; \backslash times\; 2\}\; =\; \backslash frac\{407\}\{750\}$$

3. Probability of a family having no girl (0 girls):

   Number of families with no girl = 211

   P(no girl) = $$\backslash frac\{\backslash text\{Number\; of\; families\; with\; no\; girl\}\}\{\backslash text\{Total\; number\; of\; families\}\}\; =\; \backslash frac\{211\}\{1500\}$$

   This fraction is already in its simplest form.

Checking the sum of these probabilities:

The sum of the probabilities of all possible outcomes should be 1.

Sum = P(2 girls) + P(1 girl) + P(no girl)

$$Sum\; =\; \backslash frac\{19\}\{60\}\; +\; \backslash frac\{407\}\{750\}\; +\; \backslash frac\{211\}\{1500\}$$

To add these fractions, we find a common denominator, which is 1500.

$$Sum\; =\; \backslash frac\{19\; \backslash times\; 25\}\{60\; \backslash times\; 25\}\; +\; \backslash frac\{407\; \backslash times\; 2\}\{750\; \backslash times\; 2\}\; +\; \backslash frac\{211\}\{1500\}$$

$$Sum\; =\; \backslash frac\{475\}\{1500\}\; +\; \backslash frac\{814\}\{1500\}\; +\; \backslash frac\{211\}\{1500\}$$

$$Sum\; =\; \backslash frac\{475\; +\; 814\; +\; 211\}\{1500\}\; =\; \backslash frac\{1500\}\{1500\}\; =\; 1$$

The sum of the probabilities is indeed 1, as expected.

The total number of students in the class is given as 40.

From the provided graph (or data associated with it), the number of students who were born in August is 6.

The probability of an event is the ratio of the number of favorable outcomes to the total number of possible outcomes.

Here, the favorable outcome is a student being born in August.

P(student born in August) = $$\backslash frac\{\backslash text\{Number\; of\; students\; born\; in\; August\}\}\{\backslash text\{Total\; number\; of\; students\; considered\}\}$$

$$P(\backslash text\{born\; in\; August\})\; =\; \backslash frac\{6\}\{40\}$$

Simplifying the fraction:

$$\backslash frac\{6\}\{40\}\; =\; \backslash frac\{3\; \backslash times\; 2\}\{20\; \backslash times\; 2\}\; =\; \backslash frac\{3\}\{20\}$$

Thus, the probability that a student of the class was born in August is $\backslash frac\{3\}\{20\}$.

The experiment of tossing three coins simultaneously was conducted 200 times.

The frequency of obtaining exactly 2 heads is given as 72.

The probability of an event is calculated using the experimental data as the ratio of the frequency of the event to the total number of trials.

The event of interest is getting exactly 2 heads.

P(2 heads) = $$\backslash frac\{\backslash text\{Frequency\; of\; getting\; 2\; heads\}\}\{\backslash text\{Total\; number\; of\; times\; the\; coins\; were\; tossed\}\}$$

$$P(\backslash text\{2\; heads\})\; =\; \backslash frac\{72\}\{200\}$$

Simplifying the fraction:

$$\backslash frac\{72\}\{200\}\; =\; \backslash frac\{36\; \backslash times\; 2\}\{100\; \backslash times\; 2\}\; =\; \backslash frac\{36\}\{100\}\; =\; \backslash frac\{9\; \backslash times\; 4\}\{25\; \backslash times\; 4\}\; =\; \backslash frac\{9\}\{25\}$$

Therefore, the probability of getting 2 heads when the three coins are tossed again is $\backslash frac\{9\}\{25\}$.