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

Therefore, complete numbers the outcome space 23 = 8

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) = 7/8

3. 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.

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