What is probability of getting Atmost one head when two unbiased coins are tossed?

Solution:

Given, a coin is tossed two times.

We have to find the probability of getting at most one head.

When a coin is tossed two times.

The possible outcomes are {TT, HH, TH, HT}

Number of possible outcomes = 4

Favourable outcomes = {TT, HT, TH}

Number of favourable outcomes = 3

Probability = number of favourable outcomes / number of possible outcomes

Probability of getting at most one head = 3/4

Therefore, the probability of getting at most one head is 3/4.

✦ Try This: A coin is tossed two times. Find the probability of getting at most one tail.

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 14

NCERT Exemplar Class 10 Maths Exercise 13.3 Problem 24

A coin is tossed two times. Find the probability of getting at most one head

Summary:

A coin is tossed two times. The probability of getting at most one tail is 3/4

☛ Related Questions:

• A coin is tossed 3 times. List the possible outcomes. Find the probability of getting all heads
• A coin is tossed 3 times. List the possible outcomes. Find the probability of getting at least 2 hea . . . .
• Two dice are thrown at the same time. Determine the probability that the difference of the numbers o . . . .

Here we will learn how to find the probability of tossing three coins.

Nội dung chính Show

• What is the probability of getting at most one tail if three coins are tossed?
• What is the probability of getting at most 1 tail?
• When 3 unbiased coins are tossed once what is the probability of getting at least one head?

Let us take the experiment of tossing three coins simultaneously:

When we toss three coins simultaneously then the possible of outcomes are: (HHH) or (HHT) or (HTH) or (THH) or (HTT) or (THT) or (TTH) or (TTT) respectively; where H is denoted for head and T is denoted for tail.

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) 

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.

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.

Probability

Probability

Random Experiments

Experimental Probability

Events in Probability

Empirical Probability

Coin Toss Probability

Probability of Tossing Two Coins

Probability of Tossing Three Coins

Complimentary Events

Mutually Exclusive Events

Mutually Non-Exclusive Events

Conditional Probability

Theoretical Probability

Odds and Probability

Playing Cards Probability

Probability and Playing Cards

Probability for Rolling Two Dice

Solved Probability Problems

Probability for Rolling Three Dice

9th Grade Math

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What is the probability of getting at most one tail if three coins are tossed?

Answer. Answer: probability of occurrence of atmost 1 tail =4/8 =1/2.

What is the probability of getting at most 1 tail?

The probability of getting at most one tail is 3/4.

When 3 unbiased coins are tossed once what is the probability of getting at least one head?

Ben from St Peter's followed the tree diagram and calculated out the answer: If you flip a coin three times the chance of getting at least one head is 87.5%.