Question

Determine P(E/F). Mother, Father and son lineup at random for a photograph.
E: ‘Son on one end’, F: ‘ Father in middle.

Answer 1

Let Mother-M, Father-F and Son-S.

n(S) = 3! = 6

E = {SMF, SFM, MFS, FMS},

F = {MFS, SFM}

⇒ E ∩ F = {SFM, MFS}

P(F) = \(\frac{2}{6}= \frac{1}{3}\) and P(E ∩ F) = \(\frac{2}{6}= \frac{1}{3}\)

Then, P(E/F) = \(\frac{P(E ∩ F)}{P(F)} = \frac{\frac{1}{3}}{\frac{1}{3}} = 1\)