When us toss three coins at the same time then the possible of outcomes are: (HHH) or (HHT) or (HTH) or (THH) or (HTT) or (THT) or (TTH) or (TTT) respectively; whereby H is denoted because that head and also T is denoted for tail.

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Therefore, total numbers of outcome space 23 = 8

The above explanation will help us to settle the problems on recognize the probability the tossing 3 coins.

Worked-out troubles on probability involving tossing or cram or flipping 3 coins:

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

If three coins room tossed simultaneously at random, uncover the probability of:

Solution:

Total number of trials = 250.

Number the times 3 heads appeared = 70.

Number the times 2 heads showed up = 55.

Number of times one head showed up = 75.

Number of time no head showed up = 50.

In a arbitrarily toss that 3 coins, let E1, E2, E3 and also E4 be the occasions of obtaining three heads, two heads, one head and also 0 head respectively. Then,

P(getting 3 heads) = P(E1) number of times 3 heads appeared = Total number of trials

= 70/250

= 0.28

P(getting 2 heads) = P(E2) number of times 2 heads showed up = Total number of trials

= 55/250

= 0.22

P(getting one head) = P(E3) number of times one head showed up = Total number of trials

= 75/250

= 0.30

P(getting no head) = P(E4) variety of times on head appeared = Total variety of trials

= 50/250

= 0.20

Note:

In tossing 3 coins simultaneously, the only feasible outcomes space E1, E2, E3, E4 andP(E1) + P(E2) + P(E3) + P(E4)

= (0.28 + 0.22 + 0.30 + 0.20)

= 1

2. when 3 unbiased coins space tossed once.

What is the probability of:

(iv) getting at least 1 head

(v) getting at the very least 2 heads

In tossing 3 coins, the sample an are is provided by

S = HHH, HHT, HTH, THH, HTT, THT, TTH, TTT

And, therefore, n(S) = 8.

Let E1 = occasion of getting all heads. Then,E1 = HHHand, therefore, n(E1) = 1.Therefore, P(getting all heads) = P(E1) = n(E1)/n(S) = 1/8.

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

Let E3 = event of obtaining 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) acquiring at least 1 head

Let E4 = event of obtaining at least 1 head. Then,E4 = HTT, THT, TTH, HHT, HTH, THH, HHHand, therefore, n(E4) = 7.Therefore, P(getting at the very least 1 head) = P(E4) = n(E4)/n(S) = 7/8.

(v) acquiring at the very least 2 heads

Let E5 = occasion of obtaining at the very 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.

Let E6 = event of gaining 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. three coins space tossed simultaneously 250 times and the outcomes are recorded as provided below.

If the three coins room again tossed at the same time at random, uncover the probability of getting

(ii) 2 heads and 1 tail

(iii) all tails

Solution:

(i) Total number of trials = 250.

Number of time 1 head shows up = 100.

Therefore, the probability of acquiring 1 head

= $$\frac\textrmFrequency the Favourable Trials\textrmTotal number of Trials$$

= $$\frac\textrmNumber of times 1 Head Appears\textrmTotal variety of Trials$$

= $$\frac100250$$

= $$\frac25$$

(ii) Total variety of trials = 250.

Number of time 2 heads and 1 tail appears = 64.

.

Therefore, the probability of acquiring 2 heads and also 1 tail

= $$\frac\textrmNumber of time 2 Heads and 1 attempt appears\textrmTotal number of Trials$$

= $$\frac64250$$

= $$\frac32125$$

(iii) Total number of trials = 250.

Number that times all tails appear, the is, no head appears = 38.

Therefore, the probability of getting all tails

= $$\frac\textrmNumber of times No Head Appears\textrmTotal variety of Trials$$

= $$\frac38250$$

= $$\frac19125$$.

These examples will aid us to deal with different varieties of problems based upon probability that tossing three coins.

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