What is the probability of heads in unfair coin when you flip the coin ten times? [closed]

-1

What is the probability of heads in unfair coin when you flip the coin ten times?

i came across this question and I haven't figured it out

Suppose that you flip an unfair coin ten times, where p(heads)=3/4 and p(tails)= 1/4, Find probability of

1.p(no heads)



2.p(exactly 9 heads)



3.p(exactly 7 heads)



4.p(at least 7 heads)



5.p(number of heads greater than number of tails)

i hop this help

Thanks

asked Nov 24 at 23:34

mmhh123

closed as off-topic by amWhy, Shailesh, Leucippus, KReiser, Key Flex Nov 25 at 1:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Shailesh, Leucippus, KReiser, Key Flex

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– Aditya Dua
Nov 25 at 7:18

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-1

What is the probability of heads in unfair coin when you flip the coin ten times?

i came across this question and I haven't figured it out

Suppose that you flip an unfair coin ten times, where p(heads)=3/4 and p(tails)= 1/4, Find probability of

1.p(no heads)



2.p(exactly 9 heads)



3.p(exactly 7 heads)



4.p(at least 7 heads)



5.p(number of heads greater than number of tails)

i hop this help

Thanks

asked Nov 24 at 23:34

mmhh123

closed as off-topic by amWhy, Shailesh, Leucippus, KReiser, Key Flex Nov 25 at 1:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Shailesh, Leucippus, KReiser, Key Flex

If this question can be reworded to fit the rules in the help center, please edit the question.

Please keep in mind that this is not a free homework solution service.
– Aditya Dua
Nov 25 at 7:18

add a comment |

-1

What is the probability of heads in unfair coin when you flip the coin ten times?

i came across this question and I haven't figured it out

Suppose that you flip an unfair coin ten times, where p(heads)=3/4 and p(tails)= 1/4, Find probability of

1.p(no heads)



2.p(exactly 9 heads)



3.p(exactly 7 heads)



4.p(at least 7 heads)



5.p(number of heads greater than number of tails)

i hop this help

Thanks

asked Nov 24 at 23:34

mmhh123

What is the probability of heads in unfair coin when you flip the coin ten times?

i came across this question and I haven't figured it out

Suppose that you flip an unfair coin ten times, where p(heads)=3/4 and p(tails)= 1/4, Find probability of

1.p(no heads)



2.p(exactly 9 heads)



3.p(exactly 7 heads)



4.p(at least 7 heads)



5.p(number of heads greater than number of tails)

i hop this help

Thanks

probability discrete-mathematics

asked Nov 24 at 23:34

mmhh123

asked Nov 24 at 23:34

mmhh123

asked Nov 24 at 23:34

mmhh123

asked Nov 24 at 23:34

mmhh123

asked Nov 24 at 23:34

mmhh123

closed as off-topic by amWhy, Shailesh, Leucippus, KReiser, Key Flex Nov 25 at 1:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Shailesh, Leucippus, KReiser, Key Flex

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by amWhy, Shailesh, Leucippus, KReiser, Key Flex Nov 25 at 1:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Shailesh, Leucippus, KReiser, Key Flex

If this question can be reworded to fit the rules in the help center, please edit the question.

Please keep in mind that this is not a free homework solution service.
– Aditya Dua
Nov 25 at 7:18

add a comment |

Please keep in mind that this is not a free homework solution service.
– Aditya Dua
Nov 25 at 7:18

Please keep in mind that this is not a free homework solution service.
– Aditya Dua
Nov 25 at 7:18

add a comment |

1 Answer
1

active

oldest

votes

Have a look at the binomial distribution.

Probability of no heads:
$$bigg(frac{1}{4}bigg)^{10}$$

Exactly one head:
$$binom{10}{9},bigg(frac{1}{4}bigg)bigg(frac{3}{4}bigg)^9$$

Exactly seven heads:
$$binom{10}{7},bigg(frac{1}{4}bigg)^3bigg(frac{3}{4}bigg)^7$$

At least seven heads:
$$binom{10}{7},bigg(frac{1}{4}bigg)^3bigg(frac{3}{4}bigg)^7 + binom{10}{8},bigg(frac{1}{4}bigg)^2bigg(frac{3}{4}bigg)^8 + binom{10}{9},bigg(frac{1}{4}bigg)bigg(frac{3}{4}bigg)^9 + bigg(frac{3}{4}bigg)^{10}$$

Number of heads greater than number of tails:
same idea as in the previous case but starting at 6 heads.

answered Nov 25 at 0:01

DavidPM

16618

add a comment |

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

Have a look at the binomial distribution.

Probability of no heads:
$$bigg(frac{1}{4}bigg)^{10}$$

Exactly one head:
$$binom{10}{9},bigg(frac{1}{4}bigg)bigg(frac{3}{4}bigg)^9$$

Exactly seven heads:
$$binom{10}{7},bigg(frac{1}{4}bigg)^3bigg(frac{3}{4}bigg)^7$$

At least seven heads:
$$binom{10}{7},bigg(frac{1}{4}bigg)^3bigg(frac{3}{4}bigg)^7 + binom{10}{8},bigg(frac{1}{4}bigg)^2bigg(frac{3}{4}bigg)^8 + binom{10}{9},bigg(frac{1}{4}bigg)bigg(frac{3}{4}bigg)^9 + bigg(frac{3}{4}bigg)^{10}$$

Number of heads greater than number of tails:
same idea as in the previous case but starting at 6 heads.

answered Nov 25 at 0:01

DavidPM

16618

add a comment |

Have a look at the binomial distribution.

Probability of no heads:
$$bigg(frac{1}{4}bigg)^{10}$$

Exactly one head:
$$binom{10}{9},bigg(frac{1}{4}bigg)bigg(frac{3}{4}bigg)^9$$

Exactly seven heads:
$$binom{10}{7},bigg(frac{1}{4}bigg)^3bigg(frac{3}{4}bigg)^7$$

At least seven heads:
$$binom{10}{7},bigg(frac{1}{4}bigg)^3bigg(frac{3}{4}bigg)^7 + binom{10}{8},bigg(frac{1}{4}bigg)^2bigg(frac{3}{4}bigg)^8 + binom{10}{9},bigg(frac{1}{4}bigg)bigg(frac{3}{4}bigg)^9 + bigg(frac{3}{4}bigg)^{10}$$

Number of heads greater than number of tails:
same idea as in the previous case but starting at 6 heads.

answered Nov 25 at 0:01

DavidPM

16618

add a comment |

Have a look at the binomial distribution.

Probability of no heads:
$$bigg(frac{1}{4}bigg)^{10}$$

Exactly one head:
$$binom{10}{9},bigg(frac{1}{4}bigg)bigg(frac{3}{4}bigg)^9$$

Exactly seven heads:
$$binom{10}{7},bigg(frac{1}{4}bigg)^3bigg(frac{3}{4}bigg)^7$$

At least seven heads:
$$binom{10}{7},bigg(frac{1}{4}bigg)^3bigg(frac{3}{4}bigg)^7 + binom{10}{8},bigg(frac{1}{4}bigg)^2bigg(frac{3}{4}bigg)^8 + binom{10}{9},bigg(frac{1}{4}bigg)bigg(frac{3}{4}bigg)^9 + bigg(frac{3}{4}bigg)^{10}$$

Number of heads greater than number of tails:
same idea as in the previous case but starting at 6 heads.

answered Nov 25 at 0:01

DavidPM

16618

Have a look at the binomial distribution.

Probability of no heads:
$$bigg(frac{1}{4}bigg)^{10}$$

Exactly one head:
$$binom{10}{9},bigg(frac{1}{4}bigg)bigg(frac{3}{4}bigg)^9$$

Exactly seven heads:
$$binom{10}{7},bigg(frac{1}{4}bigg)^3bigg(frac{3}{4}bigg)^7$$

At least seven heads:
$$binom{10}{7},bigg(frac{1}{4}bigg)^3bigg(frac{3}{4}bigg)^7 + binom{10}{8},bigg(frac{1}{4}bigg)^2bigg(frac{3}{4}bigg)^8 + binom{10}{9},bigg(frac{1}{4}bigg)bigg(frac{3}{4}bigg)^9 + bigg(frac{3}{4}bigg)^{10}$$

Number of heads greater than number of tails:
same idea as in the previous case but starting at 6 heads.

answered Nov 25 at 0:01

DavidPM

16618

answered Nov 25 at 0:01

DavidPM

16618

answered Nov 25 at 0:01

DavidPM

16618

answered Nov 25 at 0:01

DavidPM

16618

add a comment |

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