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**Answer**

**Two examples of a Bernolli distribution solutions**

**1. Flip a coin 10 times, count the number of heads. Here n = 10. Each flip is a trial. **

It is reasonable to assume the trials are independent. Each trial has two outcomes heads (success) and tails (failure).

The probability of success on each trial is p = 1/2 and the probability of failure is q = 1 − 1/2 = 1/2. We are interested in the variable X which counts the number of successes in 10 trials. This is an example of a Bernoulli Experiment with 10 trials

**2. A basketball player takes 2 independent free throws with a probability of 0.6 of getting a basket on each shot. Let X = the number of baskets he gets.**

**Find the probability that he misses all the shots;**

=The probability of failure=1-0.6=0.4

$=0.4\times 0.4=0.16$

**Find the probability that he gets a basket on each shot;**

$=Theprobabilityofsuccess$

$=0.6\times 0.6=0.36$

Thus the probability of getting a basket on each shot is 0.36

**Find the probability that he gets atleast one shot**

**Method i**

**$=(0.6\times 0.6)+(0.6\times 0.4)+(0.4\times 0.6)$**

**$=0.36+0.24+0.24$**

**$=0.84\phantom{\rule{0ex}{0ex}}$**

**Method ii**

To get the probability that he gets atleast one shot, we can subtract the probabbility of not getting any single shot from 1.

That is;

$=1-0.16=0.84$

=0.84

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