Question

fair coin is tossed three times. What is the probability that all three tosses land on heads,given that:(a) the first toss lands on heads(b) at least one toss lands on heads.

Answer 1

Answer:

(a) The probability that all three tosses land on heads given that the 1st toss lands on Heads is $\frac{1}{4}$.

(b) The probability that all three tosses land on heads given that at least one toss lands on heads is $\frac{1}{7}$.

Step-by-step explanation:

The sample space of tossing 3 coins is:

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} = 8 outcomes.

(a)

The sample space such that the 1st toss lands on Heads is:

S₁ = {HHH, HHT, HTH, HTT} = 4 outcomes.

The probability that all three tosses land on heads given that the 1st toss lands on Heads is:

P (3 Heads | 1st toss is Heads) = $\frac{P(3\ Heads\ \cap \ 1st\ toss\ is\ Heads)}{P(1st\ toss\ is\ Heads)}$

                                                  $=\frac{\frac{1}{4} }{\frac{4}{8} } \\=\frac{1}{8}\times \frac{8}{4}\\ =\frac{1}{4}$

Thus, the probability that all three tosses land on heads given that the 1st toss lands on Heads is $\frac{1}{4}$.

(b)

The sample space such that at least one toss lands on heads is:

S₂ = {HHH, HHT, HTH, THH, HTT, THT, TTH} = 7 outcomes

The probability that all three tosses land on heads given that at least one toss lands on heads is,

P (3 Heads | At least 1 Heads) $=\frac{P(3\ Heads\ \cap\ At\ least\ 1\ Heads)}{P( At\ least\ 1\ Heads)}$

                                                  $=\frac{\frac{1}{7} }{\frac{7}{8} } \\=\frac{1}{8}\times \frac{8}{7}\\ =\frac{1}{7}$

Thus, the probability that all three tosses land on heads given that at least one toss lands on heads is $\frac{1}{7}$.