Three coins are tossed find the probability of tossing each of the following events

In this article, we will learn how to find the probability of tossing 3 coins. We know that when a coin is tossed, the outcomes are head or tail. We can represent head by H and tail by T. Now consider an experiment of tossing three coins simultaneously. The possible outcomes will be HHH, TTT, HTT, THT, TTH, THH, HTH, HHT. So the total number of outcomes is 23 = 8. The above explanation will help us to solve the problems of finding the probability of tossing three coins.

The probability of an event E is defined as P(E) = (Number of favourable outcomes of E)/ (total number of possible outcomes of E).

Step by Step Solutions to the tossing of 3 coins Problems

The following are some problems related to the tossing of 3 coins.

Example 1. When 3 unbiased coins are tossed once.

Find the probability of:

(i) getting all tails

(ii) getting two heads

(iii) getting at least 1 head

(iv) getting one head

Solution:

When 3 coins are tossed, the possible outcomes are HHH, TTT, HTT, THT, TTH, THH, HTH, HHT.

The sample space is S = { HHH, TTT, HTT, THT, TTH, THH, HTH, HHT}

Number of elements in sample space, n(S) = 8

(i) Let E1 denotes the event of getting all tails.

E1 = {TTT}

n(E1) = 1

P(getting all tails) = n(E1)/ n(S)

= ⅛

Hence the required probability is ⅛.

(ii) Let E2 denotes the event of getting two heads.

E2 = {HHT, HTH, THH}

n(E2) = 3

P(getting two heads) = n(E2)/ n(S)

= 3/8

Hence the required probability is ⅜.

(iii) Let E3 denotes the event of getting atleast one head.

E3 = { HHH, HTT, THT, TTH, THH, HTH, HHT }

n(E3) = 7

P(getting atleast one head) = n(E3)/ n(S)

= 7/8

Hence the required probability is 7/8.

(iv) Let E4 denotes the event of getting one head.

E4 = { HTT, THT, TTH}

n(E4) = 3

P(getting one head) = n(E4)/ n(S)

= 3/8

Hence the required probability is 3/8.

Example 2: In an experiment, three coins are tossed simultaneously at random 250 times. It was found that three heads appeared 70 times, two heads appeared 55 times, one head appeared 75 times and no head appeared 50 times. find the probability of:

(i) getting three heads,

(ii) getting one head,

(iii) getting no head

(iv) getting two heads,

Solution:

The total number of trials, n(S) = 250

(i) Let E1 denotes the event of getting three heads.

n(E1) = 70

P(E1) = No. of times 3 heads appeared/ total number of trials

= n(E1) / n(S)

= 70/250

= 0.28

Hence the required probability is 0.28.

(ii) Let E2 denotes the event of getting one head.

n(E2) = 75

P(E2) = No. of times 1 head appeared/ total number of trials

= n(E2) / n(S)

= 75/250

= 0.3

Hence the required probability is 0.3.

(iii) Let E3 denote the event of getting no head.

n(E3) = 50

P(E3) = No. of times no head appeared/ total number of trials

= n(E3) / n(S)

= 50/250

= 0.2

Hence the required probability is 0.2.

(iv) Let E4 denote the event of getting 2 heads.

n(E4) = 55

P(E4) = No. of times 2 heads appeared/ total number of trials

= n(E4) / n(S)

= 55/250

= 0.22

Hence the required probability is 0.22.

Related Links:

• Bayes Theorem of Probability
• Probability Problems
• Probability JEE Main Previous Year Questions With Solutions

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