As Sandor and Martin pointed out above, it is ok if the subschemes intersect in the emptyset pairwise.  

**Example 1:**  Here's an example where it's false without that hypotheses, notice that the varieties are smooth and they intersect *pairwise* with normal crossings.  **EDIT:** as t3suji pointed out in the comments, these varieties don't intersect transversally in the ambient space, just in the ambient $z = 0$ plane **EndOfEdit**  Consider $X = \mathbb{A}^3$ and set $Z_1 = V(y,z)$, $Z_2 = V(x,z)$, $Z_3 = V(x, z-1)$.  Notice that $Z_3$ doesn't intersect any of the other subschemes.

Then, $I_1 \cap I_2 = (z, xy)$.

However, $I_1 \cdot I_2 = (xy, yz, xz, z^2)$.

These ideals are not equal clearly.  Now, we can immediately see that multiplying/intersecting by $I_3$ won't change the behavior at the origin at all since the ideal doesn't vanish there, so they are not equal.  However, just to be sure, I also did the following computation (with Macaulay2):

$$I_1 \cdot I_2 \cdot I_3 = (xz, yz, z^3 - z^2, yz^2 - yz, xz^2 - xz).$$

$$I_1 \cap I_2 \cap I_3 = (z^2 -z, xz, xy).$$

Macaulay2 also confirmed that the ideals were not equal.

**Example 2:**  Here's a different example.  a 4th variety that doesn't intersect the others at all.
$X = \mathbb{A}^3$.  $I_1 = (x,y)$, $I_2 = (x,z)$, $I_3 = (y,z)$ and $I_4 = (x-1,y-1,z-1)$.  Certainly again the $I_4$ doesn't matter, it's just included so that the sum of the ideals is equal to $R = k[x,y,z]$.  I wonder if it might be reasonable to say that $Z_1$, $Z_2$ and $Z_3$, as a triple, have transverse intersection at the origin.  Anyways:

$I_1 \cap I_2 \cap I_3 = (xy, xz, yz)$ but,
$I_1 \cdot I_2 \cdot I_3 = (yz^2, xz^2, y^2z, xyz, x^2z, xy^2, x^2y)$.

Ok, let me now give a proof of a correct statement showing that sometimes they are equal.

**Lemma:**  Suppose that subschemes $Z_1, \dots, Z_k$ have pairwise trivial intersection in some ambient Noetherian scheme $X$.  Then $I_{Z_1} \dots I_{Z_K} = I_{Z_1} \cap I_{Z_k}$.  

*Proof:*  The statement is local so we may assume that $X$ is the spectrum of a local ring $(R, \mathfrak{m})$.  Now, since $I_{Z_1} + I_{Z_2} = R$, at least one of those ideals must equal $R$ (if not, both would be in the maximal ideal $\mathfrak{m}$, and so would their sum).  Likewise with all pairs.  Therefore, at most one of the ideals $I_{Z_i}$ is not equal to $R$.  But now the statement is obvious.  $R \cdot R \dots I_{Z_i} \dots R = I_{Z_i} = R \cap R \cap \dots I_{Z_i} \cap \dots R$.