Probability of Tossing Three Coins

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

$Probability of Tossing Three Coins$

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

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

= $\frac{\textrm{Frequency of Favourable Trials}}{\textrm{Total Number of Trials}}$

= $\frac{\textrm{Number of Times 1 Head Appears}}{\textrm{Total Number of Trials}}$

= $\frac{100}{250}$

= $\frac{2}{5}$

(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

= $\frac{\textrm{Number of Times 2 Heads and 1 Trial appears}}{\textrm{Total Number of Trials}}$

= $\frac{64}{250}$

= $\frac{32}{125}$

(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

= $\frac{\textrm{Number of Times No Head Appears}}{\textrm{Total Number of Trials}}$

= $\frac{38}{250}$

= $\frac{19}{125}$.

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

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