When we toss 3 coins concurrently then the feasible of outcomes are: (HHH) or (HHT) or (HTH) or (THH) or (HTT) or (THT) or (TTH) or (TTT) respectively; wherein H is denoted for head and also T is denoted because that tail.
You are watching: Sample space of flipping a coin 3 times
The over explanation will assist us to solve the difficulties on recognize the probability of tossing three coins.
Worked-out difficulties on probability including tossing or throwing or flipping 3 coins:
1. When 3 coins space tossed randomly 250 times and also it is uncovered that 3 heads appeared 70 times, two heads appeared 55 times, one head showed up 75 times and also no head showed up 50 times.
If 3 coins space tossed concurrently at random, uncover the probability of:
(i) getting three heads,
(ii) acquiring two heads,
(iii) gaining one head,
(iv) acquiring no head
Solution:
Total variety of trials = 250.
Number that times three heads showed up = 70.
Number that times two heads appeared = 55.
Number of times one head appeared = 75.
Number of times no head appeared = 50.
In a random toss the 3 coins, allow E1, E2, E3 and E4 be the events of getting three heads, 2 heads, one head and also 0 head respectively. Then,(i) gaining three heads
P(getting three heads) = P(E1) number of times three heads appeared = Total number of trials= 70/250
= 0.28
(ii) obtaining two heads
P(getting two heads) = P(E2) number of times 2 heads showed up = 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 variety of trials= 75/250
= 0.30
(iv) acquiring no head
P(getting no head) = P(E4) number of times ~ above head showed up = Total variety of trials= 50/250
= 0.20
Note:
In tossing 3 coins simultaneously, the only feasible outcomes are E1, E2, E3, E4 andP(E1) + P(E2) + P(E3) + P(E4)= (0.28 + 0.22 + 0.30 + 0.20)
= 1
2. once 3 unbiased coins are tossed once.
What is the probability of:
(i) acquiring all heads
(ii) getting two heads
(iii) acquiring one head
(iv) gaining at the very least 1 head
(v) gaining at least 2 heads
(vi) gaining atmost 2 headsSolution:
In tossing 3 coins, the sample space is provided by
S = HHH, HHT, HTH, THH, HTT, THT, TTH, TTT
And, therefore, n(S) = 8.
(i) acquiring all heads
Let E1 = event of gaining all heads. Then,E1 = HHHand, therefore, n(E1) = 1.Therefore, P(getting all heads) = P(E1) = n(E1)/n(S) = 1/8.(ii) obtaining two heads
Let E2 = event of gaining 2 heads. Then,E2 = HHT, HTH, THHand, therefore, n(E2) = 3.Therefore, P(getting 2 heads) = P(E2) = n(E2)/n(S) = 3/8.(iii) gaining one head
Let E3 = occasion of gaining 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 the very least 1 head
Let E4 = occasion of getting at the very least 1 head. Then,E4 = HTT, THT, TTH, HHT, HTH, THH, HHHand, therefore, n(E4) = 7.Therefore, P(getting at least 1 head) = P(E4) = n(E4)/n(S) = 7/8.(v) gaining at least 2 heads
Let E5 = occasion of getting at least 2 heads. Then,E5 = HHT, HTH, THH, HHHand, therefore, n(E5) = 4.Therefore, P(getting at the very least 2 heads) = P(E5) = n(E5)/n(S) = 4/8 = 1/2.(vi) obtaining atmost 2 heads
Let E6 = occasion of acquiring atmost 2 heads. Then,E6 = HHT, HTH, HTT, THH, THT, TTH, TTTand, therefore, n(E6) = 7.Therefore, P(getting atmost 2 heads) = P(E6) = n(E6)/n(S) = 7/83. 3 coins space tossed simultaneously 250 times and the outcomes are videotaped 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 all at once at random, discover the probability that getting
(i) 1 head
(ii) 2 heads and 1 tail
(iii) every tails
Solution:
(i) Total variety of trials = 250.
Number of times 1 head shows up = 100.
Therefore, the probability of obtaining 1 head
= \(\frac\textrmFrequency of Favourable Trials\textrmTotal variety of Trials\)
= \(\frac\textrmNumber of times 1 Head Appears\textrmTotal variety of Trials\)
= \(\frac100250\)
= \(\frac25\)
(ii) Total number of trials = 250.
Number of times 2 heads and also 1 tail appears = 64.
Therefore, the probability of acquiring 2 heads and 1 tail
= \(\frac\textrmNumber of times 2 Heads and also 1 attempt appears\textrmTotal number of Trials\)
= \(\frac64250\)
= \(\frac32125\)
(iii) Total variety of trials = 250.
Number of times every tails appear, the is, no head shows up = 38.
Therefore, the probability of obtaining all tails
= \(\frac\textrmNumber of time No Head Appears\textrmTotal number of Trials\)
= \(\frac38250\)
= \(\frac19125\).
These examples will help us to resolve different types of problems based upon probability the tossing 3 coins.
See more: What Is A Flowering Seed Plant Called? Angiosperm Gymnosperm Fern Moss
Probability
Probability
Random Experiments
Experimental Probability
Events in Probability
Empirical Probability
Coin Toss Probability
Probability of Tossing two Coins
Probability that Tossing 3 Coins
Complimentary Events
Mutually to exclude, Events
Mutually Non-Exclusive Events
Conditional Probability
Theoretical Probability
Odds and also Probability
Playing Cards Probability
Probability and also Playing Cards
Probability because that Rolling two Dice
Solved Probability Problems
Probability because that Rolling 3 Dice
9th class Math
From Probability of Tossing 3 Coins to residence PAGE