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 \( P(X=3)=\frac{6}{35} \)

To find \( P(X=3) \), we need to consider the probability of Sharma opening 2 tins that are not carrots followed by 1 tin that is a carrot. 1. The total number of tins is 7 (3 carrots + 2 peas + 2 sweetcorn). 2. The probability that the first tin is not a carrot is \(\frac{4}{7}\) (2 peas + 2 sweetcorn out of 7 total). 3. The probability that the second tin opened is also not a carrot, now from 6 remaining tins, is \(\frac{3}{6} = \frac{1}{2}\). 4. Finally, the probability that the third tin is a carrot is \(\frac{3}{5}\) (3 carrots out of 5 remaining tins). Thus, we multiply these individual probabilities: \[ P(X=3) = \left(\frac{4}{7}\right) \times \left(\frac{3}{6}\right) \times \left(\frac{3}{5}\right) = \frac{4}{7} \times \frac{1}{2} \times \frac{3}{5} = \frac{12}{70} = \frac{6}{35} \] This shows the desired result!