Answer:

AaBbCcDdEE × AaBbCcDdee : This is a pentahybrid cross. To simplify it, convert it to 5 monohybrid crosses.

Aa × Aa = 1/2 (homo, AA, aa), 1/2 (hetero, Aa)

Bb × Bb = 1/2 (homo), 1/2 (hetero)

Cc × Cc = 1/2 (homo), 1/2 (hetero)

Dd × Dd = 1/2 (homo), 1/2 (hetero)

EE × ee = 1 (hetero, Ee)

We want to check probability of 3 homozygous and 2 heterozygous genes.

From the above crosses, gene E is always heterozygous. So, we will count it in heterozygous only ans take it as 1.

The other 4 genes, can be either homozygous or heterozygous. The probability of both homozygous and heterozygous is same and is equal to 1/2.

So, 3 homozygous genes = 1/2 × 1/2 × 1/2

2 heterozygous genes = 1/2 × 1 (because E is always heterozygous)

Final probability = 1/2 × 1/2 × 1/2 × 1/2 × 1 = 1/16

There are 4 such combinations (A, B, C, D genes), so final answer will be = 4 × 1/16 = 0.25