G A coin with probability 0.87 of landing heads is tossed until 2 heads are obtained. What is the probability of getting the second head on the 6th toss given that the first two tosses landed tails

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

HH

P(Heads)=0.87

Therefore P(Tails)=1-0.87=0.13

The required Probability is:

P(TTHTTH or TTTHTH or HH)

=P(TTHTTH) +P(TTTHTH)+P(HH)

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(HH) =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

the answer is (4,3)

since the rule for r0, 90º is (-y,x) and it was already applied to form the image, all you have to do is reverse the answer using this formula.