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# Make the sample space of tossing two coins simultaneously and find the probability of:

1. Exactly two heads

2. At least one head

3. At least one tail

4. At most two tails

**Given:**

Two coins are tossed simultaneously.

**To do:**

We have to find the probability of getting:

1. Exactly two heads

2. At least one head

3. At least one tail

4. At most two tails

**Solution:**

The sample space of the experiment is $(HH,\ HT,\ TH,\ TT)$.

Total number of outcomes $=\ 4$

1. Outcomes in favour of getting exactly two heads $=\ ( HH)$

Number of outcomes in favour of getting exactly two heads $=\ 1$

Probability of getting exactly heads on tossing the two coins$=\frac{Total\ number\ of\ favorable\ outcomes}{Total\ number\ of\ outcomes}$

$=\frac{1}{4}$

2. Outcomes in favour of getting at least one head on tossing the two coins $=\ ( HT,\ TH,\ HH)$

Number of outcomes in favour of getting at least one head $=\ 3$

Probability of getting at least one head on tossing the two coins$=\frac{Total\ number\ of\ favorable\ outcomes}{Total\ number\ of\ outcomes}$

$=\frac{3}{4}$

3. Outcomes in favour of getting at least one tail on tossing the two coins $=\ ( HT,\ TH,\ TT)$

Number of outcomes in favour of getting at least one tail $=\ 3$

Probability of getting at least one tail on tossing the two coins$=\frac{Total\ number\ of\ favorable\ outcomes}{Total\ number\ of\ outcomes}$

$=\frac{3}{4}$

4. Outcomes in favour of getting at most two tails on tossing the two coins $=\ ( HT,\ TH,\ TT, HH)$

Number of outcomes in favour of getting at most two tails $=\ 4$

Probability of getting at most two tails on tossing the two coins$=\frac{Total\ number\ of\ favorable\ outcomes}{Total\ number\ of\ outcomes}$

$=\frac{4}{4}$

$=1$