How To Solve: Probability Problems involving Coin Toss

Attached pdf of this Article as SPOILER at the top! Happy learning!

Hi All,

I have recently uploaded a video on YouTube to discuss Probability Problems involving Coin Toss in Detail:




Following is covered in the video

¤ What is Probability of an Event ?
¤ Probability: Tossing 1 Fair Coin
¤ Probability: Tossing 2 Fair Coins
¤ Probability: Tossing 3 Fair Coins

Theory

What is Probability of an Event?

• Probability of an Event is the Likelihood of occurrence of that event.

• Probability that an Event, E, will occur is denoted by P(E)

P(E) = "No. of successful Outcomes" /"Total number of Outcomes"


Probability: Tossing 1 Fair Coin

Fair coin is a coin which has equal probability of getting a Head or a Tail.

When we toss one coin or when we toss a coin one time then

• Total Number of Outcomes = 2 ( We can get a Head(H) or a Tail(T) )

• Outcomes are { H , T }

• Probability of Getting a Head, P(H) = \(\frac{1}{2}\) (As there are two outcomes and only one out of those results in a head)

• Probability of Getting a Tail, P(T) = \(\frac{1}{2}\) = (As there are two outcomes and only one out of those results in a Tail)


Probability: Tossing 2 Fair Coins

When we toss two coins or when we toss a coin two times then

• Total Number of Outcomes = \(2^2\) = 4 ( In each of the toss we can get a Head(H) or a Tail(T) => 2*2 = 4 )

• Outcomes are { HH, HT, TH, TT }

• Probability of Getting 0 Head, P(0H) = \(\frac{1}{4}\) (As there is ONLY one outcome out of 4 where we get 0 Head or 2 Tails)

• Probability of Getting 1 Head, P(1H) = \(\frac{2}{4}\) = \(\frac{1}{2}\) (As there are two outcomes out of 4 where we get 1 Head i.e. HT, TH)

• Probability of Getting 2 Head, P(2H) = \(\frac{1}{4}\) (As there is ONLY one outcome out of 4 where we get 2 Head i.e. HH)

• Probability of Getting 0 Head, P(0T) = \(\frac{1}{4}\) (As there is ONLY one outcome out of 4 where we get 0 Tail or 2 Heads)

• Probability of Getting 1 Head, P(1T) = \(\frac{2}{4}\) = \(\frac{1}{2}\) (As there are two outcomes out of 4 where we get 1 Tail i.e. HT, TH)

• Probability of Getting 2 Head, P(2T) = \(\frac{1}{4}\) (As there is ONLY one outcome out of 4 where we get 2 Tails i.e. TT)

• P(0H) + P(1H) + P(2H) = \(\frac{1}{4}\) + \(\frac{2}{4}\) + \(\frac{1}{4}\) = 1

• Probability of at least 1 Head = P(1H) + P(2H) = \(\frac{2}{4}\) + \(\frac{1}{4}\) = \(\frac{3}{4}\)

• Probability of at least 1 Head = 1 - P(0H) = 1 - \(\frac{1}{4}\) = \(\frac{3}{4}\)

• Probability of at least 1 Tail = 1 - P(0T) = 1 - \(\frac{1}{4}\) = \(\frac{3}{4}\)



Probability: Tossing 3 Fair Coins

When we toss three coins or when we toss a coin three times then

• Total Number of Outcomes = \(2^3\) = 8 ( In each of the toss we can get a Head(H) or a Tail(T) => 2*2*2 = 8 )

• Outcomes are { HHH, HHT, HTH, HTT, THH, THT, TTH, TTT }

• Probability of Getting 0 Head, P(0H) = \(\frac{1}{8}\) (As there is ONLY one outcome out of 8 where we get 0 Head or 3 Tails)

• Probability of Getting 1 Head, P(1H) = \(\frac{3}{8}\) (As there are three outcomes out of 8 where we get 1 Head i.e. HTT, THT, TTH)
We can also find this by finding the position of that one Head out of three slots in 3C1 = \(\frac{3!}{1!*2!}\) = 3 ways and divide it by the total number of outcomes which is 8

• Probability of Getting 2 Head, P(2H) = \(\frac{3}{8}\) (As there are three outcomes out of 8 where we get 2 Heads i.e. HHT, THH, HTH)
We can also find this by finding the position of those 2 Heads or 1 Tail out of three slots in 3C2 = \(\frac{3!}{2!*1!}\) = 3 ways and divide it by the total number of outcomes which is 8

• Probability of Getting 3 Head, P(3H) = \(\frac{1}{8}\) (As there is ONLY one outcome out of 8 where we get 3 Head i.e. HHH)

• Probability of Getting 0 Tail, P(0T) = \(\frac{1}{8}\) (As there is ONLY one outcome out of 8 where we get 0 Tail or 3 Heads)

• Probability of Getting 1 Tail, P(1T) = \(\frac{3}{8}\) (As there are three outcomes out of 8 where we get 1 Tail i.e. THH, HTH, HHT)
We can also find this by finding the position of that one Tail out of three slots in 3C1 = \(\frac{3!}{1!*2!}\) = 3 ways and divide it by the total number of outcomes which is 8

• Probability of Getting 2 Tail, P(2T) = \(\frac{3}{8}\) (As there are three outcomes out of 8 where we get 2 Tails i.e. TTH, THT, HTT)
We can also find this by finding the position of those 2 Tails or 1 Head out of three slots in 3C2 = \(\frac{3!}{2!*1!}\) = 3 ways and divide it by the total number of outcomes which is 8

• Probability of Getting 3 Tail, P(3T) = \(\frac{1}{8}\) (As there is ONLY one outcome out of 8 where we get 3 Tail i.e. TTT)

• P(0H) + P(1H) + P(2H) + P(3H) = \(\frac{1}{8}\) + \(\frac{3}{8}\) + \(\frac{3}{8}\) + \(\frac{1}{8}\) = 1

• Probability of at least 1 Head = 1 - \(\frac{1}{8}\) = \(\frac{7}{8}\)

• Probability of at least 2 Head = P(2H) + P(3H) = \(\frac{3}{8}\) + \(\frac{1}{8}\) = \(\frac{4}{8}\) = \(\frac{1}{2}\)

• Probability of at least 1 Tail = 1 - P(0T) = 1 - \(\frac{1}{8}\) = \(\frac{7}{8}\)

• Probability of at least 2 Tail = P(2T) + P(3T) = \(\frac{3}{8}\) + \(\frac{1}{8}\) = \(\frac{4}{8}\) = \(\frac{1}{2}\)











Hope it helps!
Good Luck!