1. Write the structure clearly

The compound is:

$\mathrm{CH_3-CH=CH-CH(OH)-CH_3}$

This is a five-carbon chain with:

• one double bond between C-2 and C-3,
• one hydroxyl group on C-4.

So the structure is pent-2-en-4-ol.

2. Check for geometrical isomerism about the double bond

The double bond is between C-2 and C-3:

$\mathrm{CH_3-CH=CH-CH(OH)-CH_3}$

For geometrical isomerism, each carbon of the double bond must have two different groups.

• On C-2: groups are $\mathrm{CH_3}$ and H
• On C-3: groups are $\mathrm{CH(OH)CH_3}$ and H

Since each double-bond carbon has two different substituents, E/Z (cis/trans) isomerism is possible.

Hence, number of geometrical isomers = $2$.

3. Check for optical isomerism

Now inspect the carbon bearing OH:

$\mathrm{-CH(OH)-}$

That carbon is attached to:

• $\mathrm{OH}$
• H
• $\mathrm{CH_3}$
• $\mathrm{CH=CH-CH_3}$

These are four different groups, so this carbon is a chiral center.

Thus, for each geometrical isomer, there are 2 optical isomers (an enantiomeric pair).

Hence, number of optical possibilities per geometrical form = $2$.

4. Total number of stereoisomers

Total stereoisomers:

$2 \text{ (E/Z)} \times 2 \text{ (R/S)} = 4$

5. Check for meso form

A meso form would require at least two stereocenters and internal symmetry. Here there is only one chiral center, so no meso form exists.

6. Final answer

The number of stereoisomers is:

$\boxed{4}$

So the correct option is B.