Probability of Tossing Three Coins

Here we will learn how to find the probability of tossing three coins.

Let us take the experiment of tossing three coins simultaneously:

When we toss three coins simultaneously then the possible of outcomes are: (HHH) or (HHT) or (HTH) or (THH) or (HTT) or (THT) or (TTH) or (TTT) respectively; where H is denoted for head and T is denoted for tail.

Therefore, total numbers of outcome are 23 = 8

The above explanation will help us to solve the problems on finding the probability of tossing three coins.

Worked-out problems on probability involving tossing or throwing or flipping three coins:

1. When 3 coins are tossed randomly 250 times and it is found that three heads appeared 70 times, two heads appeared 55 times, one head appeared 75 times and no head appeared 50 times.

If three coins are tossed simultaneously at random, find the probability of:
(i) getting three heads,
(ii) getting two heads,
(iii) getting one head,
(iv) getting no head

Solution:

Total number of trials = 250.
Number of times three heads appeared = 70.
Number of times two heads appeared = 55.
Number of times one head appeared = 75.
Number of times no head appeared = 50.

In a random toss of 3 coins, let E1, E2, E3 and E4 be the events of getting three heads, two heads, one head and 0 head respectively. Then,

(i) getting three heads

P(getting three heads) = P(E1)

      Number of times three heads appeared
=                   Total number of trials         

= 70/250
= 0.28


(ii) getting two heads

P(getting two heads) = P(E2)

      Number of times two heads appeared
=                 Total number of trials         

= 55/250
= 0.22


(iii) getting one head

P(getting one head) = P(E3)

      Number of times one head appeared
=                 Total number of trials        

= 75/250
= 0.30


(iv) getting no head

P(getting no head) = P(E4)

      Number of times on head appeared
=                 Total number of trials      

= 50/250
= 0.20


Note:

In tossing 3 coins simultaneously, the only possible outcomes are E1, E2, E3, E4 and P(E1) + P(E2) + P(E3) + P(E4)

= (0.28 + 0.22 + 0.30 + 0.20)
= 1


2. When 3 unbiased coins are tossed once.

What is the probability of:
(i) getting all heads
(ii) getting two heads
(iii) getting one head
(iv) getting at least 1 head
(v) getting at least 2 heads
(vi) getting atmost 2 heads

Solution:

In tossing three coins, the sample space is given by
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
And, therefore, n(S) = 8.

(i) getting all heads

Let E1 = event of getting all heads. Then,
E1 = {HHH}
and, therefore, n(E1) = 1.
Therefore, P(getting all heads) = P(E1) = n(E1)/n(S) = 1/8.

(ii) getting two heads

Let E2 = event of getting 2 heads. Then,
E2 = {HHT, HTH, THH}
and, therefore, n(E2) = 3.
Therefore, P(getting 2 heads) = P(E2) = n(E2)/n(S) = 3/8.

(iii) getting one head

Let E3 = event of getting 1 head. Then,
E3 = {HTT, THT, TTH} and, therefore,
n(E3) = 3.
Therefore, P(getting 1 head) = P(E3) = n(E3)/n(S) = 3/8.

(iv) getting at least 1 head

Let E4 = event of getting at least 1 head. Then,
E4 = {HTT, THT, TTH, HHT, HTH, THH, HHH}
and, therefore, n(E4) = 7.
Therefore, P(getting at least 1 head) = P(E4) = n(E4)/n(S) = 7/8.

(v) getting at least 2 heads

Let E5 = event of getting at least 2 heads. Then,
E5 = {HHT, HTH, THH, HHH}
and, therefore, n(E5) = 4.
Therefore, P(getting at least 2 heads) = P(E5) = n(E5)/n(S) = 4/8 = 1/2.

(vi) getting atmost 2 heads

Let E6 = event of getting atmost 2 heads. Then,
E6 = {HHT, HTH, HTT, THH, THT, TTH, TTT}
and, therefore, n(E6) = 7.
Therefore, P(getting atmost 2 heads) = P(E6) = n(E6)/n(S) = 7/8

These examples will help us to solve different types of problems based on probability of tossing three coins.

9 Grade Math

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