CBSE Class 9 Mathematics Chapter 15 Statistics 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 probability that she did not hit a boundary, we first need to determine the number of balls on which she did not hit a boundary.

Number of balls where no boundary was hit = Total balls played - Number of boundaries hit

= 30 - 6 = 24 balls.

The empirical probability of an event is calculated as:

$$P(\backslash text\{Event\})\; =\; \backslash frac\{\backslash text\{Number\; of\; favorable\; outcomes\}\}\{\backslash text\{Total\; number\; of\; trials\}\}$$

In this case, the event is 'not hitting a boundary'.

So, the probability that she did not hit a boundary is:

$$P(\backslash text\{did\; not\; hit\; a\; boundary\})\; =\; \backslash frac\{\backslash text\{Number\; of\; balls\; in\; which\; she\; did\; not\; hit\; a\; boundary\}\}\{\backslash text\{Total\; number\; of\; balls\; played\}\}\; =\; \backslash frac\{24\}\{30\}$$

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.

The data recorded is as follows:

• Number of families with 0 girls = 211
• Number of families with 1 girl = 814
• Number of families with 2 girls = 475

We will compute the probability for each case using the formula: $P(\backslash text\{Event\})\; =\; \backslash frac\{\backslash text\{Number\; of\; favorable\; outcomes\}\}\{\backslash text\{Total\; number\; of\; trials\}\}$

(i) Probability of a family having 2 girls:

Number of families with 2 girls = 475

Total number of families = 1500

$$P(\backslash text\{2\; girls\})\; =\; \backslash frac\{475\}\{1500\}$$

To simplify the fraction, we can divide both numerator and denominator by their greatest common divisor. Both are divisible by 25:

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

(ii) Probability of a family having 1 girl:

Number of families with 1 girl = 814

Total number of families = 1500

$$P(\backslash text\{1\; girl\})\; =\; \backslash frac\{814\}\{1500\}$$

Both numbers are divisible by 2:

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

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

Number of families with 0 girls = 211

Total number of families = 1500

$$P(\backslash text\{0\; girls\})\; =\; \backslash frac\{211\}\{1500\}$$

This fraction cannot be simplified further as 211 is a prime number and not a factor of 1500.

Checking the sum of these probabilities:

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

$$P(\backslash text\{2\; girls\})\; +\; P(\backslash text\{1\; girl\})\; +\; P(\backslash text\{0\; girls\})\; =\; \backslash frac\{19\}\{60\}\; +\; \backslash frac\{407\}\{750\}\; +\; \backslash frac\{211\}\{1500\}$$

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

• $\backslash frac\{19\}\{60\}\; =\; \backslash frac\{19\; \backslash times\; 25\}\{60\; \backslash times\; 25\}\; =\; \backslash frac\{475\}\{1500\}$
• $\backslash frac\{407\}\{750\}\; =\; \backslash frac\{407\; \backslash times\; 2\}\{750\; \backslash times\; 2\}\; =\; \backslash frac\{814\}\{1500\}$
• $\backslash frac\{211\}\{1500\}$ (already has the common denominator)

Now, add the numerators:

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

The sum of the probabilities is indeed 1, which confirms that these three outcomes cover all possibilities for a family with 2 children.

The total number of students in the section of Class IX is given as 40. This represents the total number of trials.

From the provided graph (or data), we can find the number of students who were born in August.

Number of students born in August = 6.

The probability of an event is calculated as:

$$P(\backslash text\{Event\})\; =\; \backslash frac\{\backslash text\{Number\; of\; favorable\; outcomes\}\}\{\backslash text\{Total\; number\; of\; trials\}\}$$

Here, the event is 'a student was born in August'.

So, the probability that a student was born in August is:

$$P(\backslash text\{born\; in\; August\})\; =\; \backslash frac\{\backslash text\{Number\; of\; students\; born\; in\; August\}\}\{\backslash text\{Total\; number\; of\; students\; considered\}\}\; =\; \backslash frac\{6\}\{40\}$$

Simplifying the fraction:

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

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

The experiment consists of tossing three coins simultaneously 200 times. This means the total number of trials is 200.

The frequencies of the different outcomes are given:

• Number of times 3 heads occurred = 23
• Number of times 2 heads occurred = 72
• Number of times 1 head occurred = 28
• Number of times 0 heads occurred = 77

We need to compute the probability of getting 2 heads when the three coins are tossed again. This is an empirical probability based on the given data.

The probability of an event is calculated as:

$$P(\backslash text\{Event\})\; =\; \backslash frac\{\backslash text\{Number\; of\; favorable\; outcomes\}\}\{\backslash text\{Total\; number\; of\; trials\}\}$$

In this case, the event is 'getting 2 heads'.

The number of times 2 heads occurred is 72.

The total number of times the coins were tossed is 200.

So, the probability of getting 2 heads is:

$$P(\backslash text\{2\; heads\})\; =\; \backslash frac\{\backslash text\{Frequency\; of\; 2\; heads\}\}\{\backslash text\{Total\; number\; of\; tosses\}\}\; =\; \backslash frac\{72\}\{200\}$$

To simplify the fraction, we can divide both the numerator and the denominator by their greatest common divisor. Both are divisible by 8:

$$\backslash frac\{72\; \backslash div\; 8\}\{200\; \backslash div\; 8\}\; =\; \backslash frac\{9\}\{25\}$$

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