Respuesta :
Answer: B. 135/364
In solving this kind of probability problems, we have to use the Hyper geometric Probability without replacement. We are given with two groups which would be the source of the 4 members of the new group, 2 coming from each respective group.
Equation:
P = [(pCr)(qCm)]/(sCt)
= [(10C2)(6C2)]/(16C4)
= 135/364
In solving this kind of probability problems, we have to use the Hyper geometric Probability without replacement. We are given with two groups which would be the source of the 4 members of the new group, 2 coming from each respective group.
Equation:
P = [(pCr)(qCm)]/(sCt)
= [(10C2)(6C2)]/(16C4)
= 135/364