probability of 3 coins real life example
Answers
Step-by-step explanation:
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.
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
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
3. Three coins are tossed simultaneously 250 times and the outcomes are recorded as given below.
Outcomes
3 heads
2 heads
1 head
No head
Total
Frequencies
48
64
100
38
250
If the three coins are again tossed simultaneously at random, find the probability of getting
(i) 1 head
(ii) 2 heads and 1 tail
(iii) All tails
Solution:
(i) Total number of trials = 250.
Number of times 1 head appears = 100.
Therefore, the probability of getting 1 head
= Frequency of Favourable TrialsTotal Number of Trials
= Number of Times 1 Head AppearsTotal Number of Trials
= 100250
= 25
(ii) Total number of trials = 250.
Number of times 2 heads and 1 tail appears = 64.
[Since, three coins are tossed. So, when there are 2 heads, there will be 1 tail also].
Therefore, the probability of getting 2 heads and 1 tail
= Number of Times 2 Heads and 1 Trial appearsTotal Number of Trials
= 64250
= 32125
(iii) Total number of trials = 250.
Number of times all tails appear, that is, no head appears = 38.
Therefore, the probability of getting all tails
= Number of Times No Head AppearsTotal Number of Trials
= 38250
= 19125.
These examples will help us to solve different types of problems based on probability of tossing three coins.