Question

A coin is tossed twice if the coin shows head it is tossed again but if it shows a tail then a die is tossed. If 8 possible outcomes are equally likely. Find the probability that the die shows a number greater than 4, if it is known that the first throw of the coin results in a tail.

Answer 1

The options for this question are missing, here are the options:

A) 1/3

B) 2/3

C) 2/5

D) 4/15

Answer:

Sample Space S = { HH, HT, T1, T2, T3, T4, T5, T6 }

We will let X be the event which shows that the number is greater than 4 and Y be the event where the first throw of the coin will result in a tail:

A = { T5, T6 }  

B = { T1, T2, T3, T4, T5, T6 }

So the probability for this is:

P(A/B)

=P(A∩B)/P(B)

=2/8/6/8/

= 1/3

Answer 2

Answer:

0.33

Step-by-step explanation:

We are given that a coin is tossed twice if the coin shows head it is tossed again but it it shows a tail then a die is tossed

Sample space={HH,HT,(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)}

Total number of outcomes=8

Let A be even that die shows a number greater than 4

A={(T,5),(T,6)}

B be event that first thrown of the coin results in a tail

B={(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)}

$A\cap B={(T,5),(T,6)}$

$P(A)=\frac{2}{8}=\frac{1}{4}$

$P(B)=\frac{6}{8}=\frac{3}{4}$

$P(A\cap B)=\frac{2}{8}=\frac{1}{4}$

We have to find the probability

$P(\frac{A}{B})=\frac{P(A\cap B)}{P(B)}$

$P(A/B)=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}=0.33$

Hence, the probability that the die shows a number greater than 4,if it known that the first throw of the coin results in a tail=0.33