CBSE Class 11 Maths Exemplar Chapter 16 Probability Solutions

The word 'ALGORITHM' has 9 distinct letters. We want to find the probability that the letters 'GOR' stay together when arranged randomly.

First, let's consider the letters 'GOR' as a single block or unit. Now, we have the following units to arrange: (GOR), A, L, I, T, H, M. This gives us a total of 7 units.

The number of ways to arrange these 7 units is $7!$.

Within the block 'GOR', the letters can also be arranged among themselves. The number of ways to arrange the letters G, O, R is $3!$.

So, the total number of arrangements where 'GOR' remain together is $7!\; \backslash times\; 3!$.

The total number of possible arrangements of the 9 distinct letters in 'ALGORITHM' is $9!$.

The required probability is the ratio of the number of favorable arrangements to the total number of arrangements:

$$P(\backslash text\{\text{'}GOR\text{'}\; together\})\; =\; \backslash frac\{\backslash text\{Number\; of\; arrangements\; with\; \text{'}GOR\text{'}\; together\}\}\{\backslash text\{Total\; number\; of\; arrangements\}\}$$

$$P(\backslash text\{\text{'}GOR\text{'}\; together\})\; =\; \backslash frac\{7!\; \backslash times\; 3!\}\{9!\}\; =\; \backslash frac\{7!\; \backslash times\; 3!\}\{9\; \backslash times\; 8\; \backslash times\; 7!\}\; =\; \backslash frac\{3!\}\{9\; \backslash times\; 8\}\; =\; \backslash frac\{6\}\{72\}\; =\; \backslash frac\{1\}\{12\}$$

Answer: The probability that the letters 'GOR' must remain together as a unit is $\backslash frac\{1\}\{12\}$.

There are 6 new employees to be assigned to 6 desks in a row. The total number of ways to assign these 6 employees to the 6 desks is $6!$.

Let's first calculate the probability that the married couple *will* have adjacent desks.

Consider the married couple as a single unit. Now we have 5 units to arrange: (Couple), Employee 3, Employee 4, Employee 5, Employee 6. The number of ways to arrange these 5 units is $5!$.

Within the couple unit, the two individuals can swap places. So, there are $2!$ ways to arrange the couple internally.

The number of arrangements where the married couple sits together is $5!\; \backslash times\; 2!$.

The probability that the married couple sits in adjacent desks is:

$$P(\backslash text\{Couple\; adjacent\})\; =\; \backslash frac\{\backslash text\{Number\; of\; arrangements\; with\; couple\; adjacent\}\}\{\backslash text\{Total\; number\; of\; arrangements\}\}$$

$$P(\backslash text\{Couple\; adjacent\})\; =\; \backslash frac\{5!\; \backslash times\; 2!\}\{6!\}\; =\; \backslash frac\{5!\; \backslash times\; 2\}\{6\; \backslash times\; 5!\}\; =\; \backslash frac\{2\}\{6\}\; =\; \backslash frac\{1\}\{3\}$$

We are asked for the probability that the married couple will have *non-adjacent* desks. This is the complement of the event that they have adjacent desks.

$$P(\backslash text\{Couple\; non-adjacent\})\; =\; 1\; -\; P(\backslash text\{Couple\; adjacent\})$$

$$P(\backslash text\{Couple\; non-adjacent\})\; =\; 1\; -\; \backslash frac\{1\}\{3\}\; =\; \backslash frac\{2\}\{3\}$$

Answer: The probability that the married couple will have non-adjacent desks is $\backslash frac\{2\}\{3\}$.

We need to find the probability that a randomly chosen integer between 1 and 1000 (inclusive) is a multiple of 2 or a multiple of 9. Let S be the sample space, so $n(S)\; =\; 1000$.

Let A be the event that the integer is a multiple of 2.

The multiples of 2 are 2, 4, 6, ..., 1000. The number of multiples of 2 is $n(A)\; =\; \backslash frac\{1000\}\{2\}\; =\; 500$.

Let B be the event that the integer is a multiple of 9.

The multiples of 9 are 9, 18, 27, ..., 999. To find the number of multiples of 9, we divide 1000 by 9: $\backslash lfloor\; \backslash frac\{1000\}\{9\}\; \backslash rfloor\; =\; 111$. So, $n(B)\; =\; 111$.

We are interested in the event A or B, which is $A\; \backslash cup\; B$. The formula for the probability of the union of two events is $P(A\; \backslash cup\; B)\; =\; P(A)\; +\; P(B)\; -\; P(A\; \backslash cap\; B)$.

$A\; \backslash cap\; B$ is the event that the integer is a multiple of both 2 and 9, which means it is a multiple of their least common multiple, LCM(2, 9) = 18.

The multiples of 18 are 18, 36, 54, ..., 990. To find the number of multiples of 18, we divide 1000 by 18: $\backslash lfloor\; \backslash frac\{1000\}\{18\}\; \backslash rfloor\; =\; 55$. So, $n(A\; \backslash cap\; B)\; =\; 55$.

Now we can find the number of integers that are multiples of 2 or 9:

$$n(A\; \backslash cup\; B)\; =\; n(A)\; +\; n(B)\; -\; n(A\; \backslash cap\; B)$$

$$n(A\; \backslash cup\; B)\; =\; 500\; +\; 111\; -\; 55\; =\; 556$$

The probability is the number of favorable outcomes divided by the total number of outcomes:

$$P(A\; \backslash cup\; B)\; =\; \backslash frac\{n(A\; \backslash cup\; B)\}\{n(S)\}\; =\; \backslash frac\{556\}\{1000\}$$

$$P(A\; \backslash cup\; B)\; =\; 0.556$$

Answer: The probability that the integer is a multiple of 2 or a multiple of 9 is $0.556$.

The experiment involves rolling a standard six-sided die repeatedly until the outcome '2' appears.

Part (i): The 2 appears on the kth roll.

For the outcome '2' to appear specifically on the kth roll, the following conditions must be met:

• The first $(k-1)$ rolls must *not* result in a '2'. There are 5 possible outcomes (1, 3, 4, 5, 6) for each of these rolls.
• The kth roll *must* result in a '2'. There is only 1 possible outcome for this roll.

The number of elements in the sample space corresponding to this event is the product of the number of possibilities for each roll:

$$\backslash text\{Number\; of\; elements\}\; =\; (\backslash text\{Number\; of\; outcomes\; not\; 2\})^\{k-1\}\; \backslash times\; (\backslash text\{Number\; of\; outcomes\; equal\; to\; 2\})$$

$$\backslash text\{Number\; of\; elements\}\; =\; 5^\{k-1\}\; \backslash times\; 1\; =\; 5^\{k-1\}$$

Part (ii): The 2 appears not later than the kth roll.

This means the outcome '2' can appear on the 1st roll, or the 2nd roll, ..., or the kth roll.

• Case 1: '2' appears on the 1st roll. There is 1 way for this to happen.
• Case 2: '2' appears on the 2nd roll. This means the 1st roll is not '2' (5 outcomes), and the 2nd roll is '2' (1 outcome). Number of ways = $5\; \backslash times\; 1\; =\; 5$.
• Case 3: '2' appears on the 3rd roll. This means the 1st and 2nd rolls are not '2' (5 outcomes each), and the 3rd roll is '2' (1 outcome). Number of ways = $5\; \backslash times\; 5\; \backslash times\; 1\; =\; 5^2$.
• ...
• Case k: '2' appears on the kth roll. As calculated in Part (i), the number of ways is $5^\{k-1\}$.

The total number of elements for the event '2 appears not later than the kth roll' is the sum of the elements from each case:

$$\backslash text\{Total\; elements\}\; =\; 1\; +\; 5\; +\; 5^2\; +\; \backslash dots\; +\; 5^\{k-1\}$$

This is a geometric series with the first term $a\; =\; 1$, common ratio $r\; =\; 5$, and $k$ terms.

The sum of a geometric series is given by $S\_k\; =\; \backslash frac\{a(r^k\; -\; 1)\}\{r\; -\; 1\}$.

$$\backslash text\{Total\; elements\}\; =\; \backslash frac\{1(5^k\; -\; 1)\}\{5\; -\; 1\}\; =\; \backslash frac\{5^k\; -\; 1\}\{4\}$$

Answer:

• (i) The number of elements for the event that '2' appears on the kth roll is $5^\{k-1\}$.
• (ii) The number of elements for the event that '2' appears not later than the kth roll is $\backslash frac\{5^k\; -\; 1\}\{4\}$.