Question

In a group of 50 scouts in a camp, 30 are well trained in first aid, remaining are well trained in hospitality and not first aid. 2 scouts are selected at random.Find the probability distribution of the no. Of selected scouts who are well trained in first aid.Find the mean of the distribution.Write one more value which is expected from a scout

Answer 1

hey friend your answer is

In a group of 50 scouts in a camp, 30 are well trained in first aid, remaining are well trained in hospitality and not first aid. 2 scouts are selected at random.Find the probability distribution of the no. Of selected scouts who are well trained in first aid.Find the mean of the distribution.Write one more value which is expected from a scout

50 x 0.6 = 3

Answer 2

the mean number of selected scouts who are well trained in first aid is approximately 1.204.

One more value that is expected from a scout is their skill level in any other relevant areas such as camping, hiking, or survival skills. Scouts are typically expected to have a well-rounded set of skills to be able to contribute effectively to their team and complete their tasks in a camp.

There are 50 scouts in total at the camp, 30 of them have received first aid training, while the other 20 have received hospitality training only.

There are three conceivable outcomes if two scouts are chosen at random:

Both of the chosen scouts have first aid training.

One of the chosen scouts has first aid training, and the other has hospitality training.

Both of the chosen scouts have a strong background in hospitality.

Let X be the number of selected scouts who are well trained in first aid. Then, the probability distribution of X is given by:

P(X=0) = P(both selected scouts are well trained in hospitality) = $(20/50) * (19/49) = 0.153$

P(X=1) = P(one selected scout is well trained in first aid and the other is well trained in hospitality) =$2 * (30/50) * (20/49) = 0.490$

P(X=2) = P(both selected scouts are well trained in first aid) = $(30/50) * (29/49) = 0.357$

The mean of this distribution is given by:

$Mean = E(X) = Σ [x * P(X=x)] for x = 0, 1, 2\\Mean = 0 * 0.153 + 1 * 0.490 + 2 * 0.357 = 1.204$

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