Sharma knows that the has 3 tits of carroks 2 tins of peas and 2 tins of swectoom ia her caplooud All the linc are the came chape and sias. but the labeis have all oeen renooned. wo Sinarmu diect ons know whal each tin contains. Sharma wants carrots for her meal. and she starts opening the tins one at a time. choscn randemly. until she opens a tin of carrots. The random variable X is the number of tins that she needs to open (2) (a) Show that $$ P(X=3)=\frac{6}{35} $$

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To find $$ P(X=3) $$, we need to determine the probability that Sharma opens exactly 3 tins before finding a tin of carrots. This means that the first two tins she opens must not contain carrots, and the third tin must contain carrots.

### Known Conditions:
- Total tins: 3 tins of carrots, 2 tins of peas, and 2 tins of sweetcorn.
- Total number of tins = $$ 3 + 2 + 2 = 7 $$.

### Step-by-Step Calculation:

1. **Determine the total number of ways to choose 3 tins from 7:**
   The total number of ways to choose 3 tins from 7 is given by the combination formula:
   $$
   \binom{n}{k} = \frac{n!}{k!(n-k)!}
   $$
   where $$ n $$ is the total number of items, and $$ k $$ is the number of items to choose.

Here, $$ n = 7 $$ and $$ k = 3 $$:
   $$
   \binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35
   $$

2. **Determine the number of favorable outcomes for $$ P(X=3) $$:**
   For $$ P(X=3) $$, the first two tins must not contain carrots, and the third tin must contain carrots.

- The number of ways to choose 2 tins from the 4 tins that are not carrots (2 peas + 2 sweetcorn):
   $$
   \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6
   $$

- The third tin must be one of the 3 tins of carrots, which can be chosen in 1 way.

Therefore, the total number of favorable outcomes for $$ P(X=3) $$ is:
   $$
   \text{Favorable outcomes} = \binom{4}{2} \times 1 = 6
   $$

3. **Calculate the probability $$ P(X=3) $$:**
   The probability is given by the ratio of the number of favorable outcomes to the total number of outcomes:
   $$
   P(X=3) = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{6}{35}
   $$

Thus, we have shown that:
$$
P(X=3) = \frac{6}{35}
$$
The probability that Sharma opens exactly 3 tins before finding a tin of carrots is $$ \frac{6}{35} $$.

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