Answer:
0.0006485 (Correct to 4 s.f)
Step-by-step explanation:
If the first two tosses landed tails, and we are concerned with getting the second head on the 6th toss.
TT***H
Since the first Head can be in any if the three empty toss above The probable sample space is:
TTHTTH
TTTHTH
TTTTHH
P(Heads)=0.87
Therefore P(Tails)=1-0.87=0.13
The required Probability is:
P(TTHTTH or TTTHTH or TTTTHH)
=P(TTHTTH) +P(TTTHTH)+P(TTTTHH)
In each case, we have 4 Tails and 2 Heads, so the Probabilities will be the same.
P(TTHTTH)=(0.13 X 0.13 X 0.87 X 0.13 X 0.13 X 0.87)= 0.00021617821
Similarly,
P(TTTHTH) =0.00021617821
P(TTTTHH) =0.00021617821
P(probability of getting the second head on the 6th toss given that the first two tosses landed tails)
=0.00021617821+0.00021617821+0.00021617821
=0.00064853463
Since no degree of accuracy is given, I'll round it up to 4 significant figures.
=0.0006485
