Question

Suppose that a biased coin that lands on heads with probability $p$is flipped $10$times. Given that a total of $6$heads results, find the conditional probability that the first $3$outcomes are

(a) $h,t,t$(meaning that the first flip results in heads, the second is tails, and the third in tails);

(b)$t,h,t.$

Short answer

The conditional probability that the first $3$outcomes are $h,t,t$is $\frac{1}{10}.$

The conditional probability that the first $3$outcomes are $t,h,t$is localid="1646454608584" $\frac{1}{10}.$

Step-by-step solution

Given Information (Part-a)

From the information, observe that a biased coin that lands on heads with probability $p$is flipped 10 times.

Number of times a coin is tossed,$n=10$

Number heads,$h=6$

A biased coin that lands on heads with probability is$P.$

Let $X$denote getting outcome.

Here, the random variable $X$follows a binomial distribution with probability of success $p$and the number of trials $10$.

The probability mass function of binomial distribution can be defined as,

$P(X=h)=\left(10C_h\right)p^h(1-p{)}^{10-h}$

Solution of the Problem (Part-a)

Calculate the conditional probability that the first $3$outcomes are $h,t,t.$

That is find role="math" localid="1646455807425" $P(h,t,t\backslash 6h).$

$P(h,t,t\mid 6h)=\frac{P(h,t,t\cap 6h)}{P\left(6h\right)}$

$=\frac{P(h,t,t)P\left(5\text{heads occur in the last}7\text{trials}\right)}{P\left(6h\right)}$

$=\frac{p\times (1-p)\times (1-p)\times \left(7C_5\right)p^5(1-p{)}^2}{\left(10C_6\right)p^6(1-p{)}^4}$

We get,

$=\frac{7C_5}{10C_6}$

$=\frac{1}{10}$

Final Answer (Part-a)

Therefore, the conditional probability that the first 3 outcomes are $h,t,t$ is$\frac{1}{10}.$

Given Information (Part-b)

Observing that a biased coin that lands on heads with probability is flipped times from the information is flipped $10$times.

Number of times a coin is tossed,$n=10$

Number heads, $h=6$

A biased coin that lands on heads with probability is $p.$

Let $X$denotes getting outcome.

Here, the random variable $X$follows a binomial distribution with probability of success $p$and the number of trials $10.$

The probability mass function of binomial distribution can be defined as,

$P(X=h)=\left(10C_h\right)p^h(1-p{)}^{10-h}$

Solution of the problem (Part-b)

Find the conditional probability that the first $3$outcomes are$t,h,t.$

That is find $P(t,h,t\mid 6h)$

$P(t,h,t\mid 6h)=\frac{P(t,h,t\cap 6h)}{P\left(6h\right)}$

$=\frac{P(t,h,t)\times P\left(5\text{heads occur in the last}7\text{trials}\right)}{P\left(6h\right)}$

$=\frac{(1-p)\times p\times (1-p)\times \left(7C_5\right)p^3(1-p{)}^2}{\left(10C_6\right)p^6(1-p{)}^4}$

We get,

$=\frac{7C_5}{10C_6}$

$=\frac{1}{10}$

Final Answer (Part-b)

Therefore, the conditional probability that the first$3$outcomes are $t,h,t$ is$\frac{1}{10}.$