# What is the probability of obtain tails and an odd number?

Question
What is the probability of obtain tails and an odd number?

2020-10-21
The coin has 2 possible outcomes: heads H and tails T/ The die has 6 possible outcomes: 1, 2, 3, 4, 5, 6.
There are then $$\displaystyle{2}\cdot{6}={12}$$ possible outcomes of flipping the coin and rolling the die: H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6.
# of possible outcomes = 12
We note that 3 of the 12 possible outcomes result in tails and an odd number: T1, 73, T5.
# of favorable outcomes = 3
The probability is the number of favorable outcomes divided by the number of possible outcomes:
P(tails and odd) = # of favorable outcomes /# of possible outcomes Te hie outcomes > $$\displaystyle=\frac{{3}}{{12}}=\frac{{1}}{{4}}={0.25}={25}\%$$

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The problem reads: Suppose $$\displaystyle{P}{\left({X}_{{1}}\right)}={.75}$$ and $$\displaystyle{P}{\left({Y}_{{2}}{\mid}{X}_{{1}}\right)}={.40}$$. What is the joint probability of $$\displaystyle{X}_{{1}}$$ and $$\displaystyle{Y}_{{2}}$$?
This is how I answered it. P($$\displaystyle{X}_{{1}}$$ and $$\displaystyle{Y}_{{2}}$$) $$\displaystyle={P}{\left({X}_{{1}}\right)}\times{P}{\left({Y}_{{1}}{\mid}{X}_{{1}}\right)}={.75}\times{.40}={0.3}.$$
What I don't understand is how do you get the $$\displaystyle{P}{\left({Y}_{{1}}{\mid}{X}_{{1}}\right)}$$? I am totally new to Statistices and I need to understand each part of the process in order to get the whole concept. Can anyone help me to understand why the P and X exist and what they represent?