Three unbiased coins are tossed what is the probability of getting at most two heads

`1/8` `7/8` `3/8``1/4`

Answer : B

Solution : When 3 coins are tossed simultaneously, all possible outcomes are HHH, HHT, HTH, THH, HTT, THT, TTH, TTT. <br> Total number of possible outcomes = 8. <br> (i) Let `E_(1)` be the event of getting exactly 2 heads. <br> Then, the favourable outcomes are HHT, HTH, THH. <br> Number of favourable outcomes = 3. <br> ` :. ` P(getting exactly 2 heads) = ` P(E_(1)) = 3/8`. <br> (ii) Let `E_(2)` be the event of getting at least 2 heads. <br> Then, `E_(2)` is the event of getting 2 or 3 heads. <br> So, the favourable outcomes are <br> HHT, HTH, THH, HHH. <br> Number of favourable outcomes = 4. <br> `:. ` P(getting at least 2 heads) = `P(E_(2)) = 4/8 = 1/2`. <br> (iii) Let `E_(3)` be the event of getting at most 2 heads. <br> Then, `E_(3)` is the event of getting 0 or 1 head or 2 heads. <br> So, the favourable outcomes are <br> TTT, HTT, THT, TTH, HHT, HTH, THH. <br> Number of favourable outcomes = 7. <br> `:. ` P(getting at most 2 heads ) = `P(E_(3)) = 7/8`.

Answer

Verified

Hint : When we toss an unbiased coin we get either a head or tail. Here we are throwing three such coins. We are going to proceed with the thought of getting head and tail sequence on three coins.

Elementary events associated with the random experiment of tossing three coins are HHH, HHT, HTH, THH, HTT, THT, TTH and TTT.
Total number of elementary events = 8
If any of the elementary events HHH, HHT, HTH and THH is an outcome, then we say that the event of “Getting at least two heads” occurs.
$\therefore $ Number of favorable elementary events = 4
$ \Rightarrow P(E) = \frac{{n(E)}}{{n(S)}}$
Hence required probability = $\frac{4}{8} = \frac{1}{2}$

Note:
Probability of an event E is (Number of favorable outcomes to event E) / (Total number of possible outcomes in sample space)
$ \Rightarrow P(E) = \frac{{n(E)}}{{n(S)}}$
In the given problem our event is getting at least two heads while tossing three coins.

Three unbiased coins are tossed. What is the probability of getting at most two heads ?

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• An unbiased coin is tossed three times. What is the probability of getting at least two heads?
• Answer (Detailed Solution Below)
• What is the probability of getting 2 heads when 3 unbiased coins are tossed?
• When 3 unbiased coins are tossed once what is the probability of getting all heads?
• When a unbiased coin is tossed thrice Where is the probability of getting three heads?

A. 3/4

B. 1/4

C. 3/8

D. 7/8

Here, S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}
Let E = event of getting at most two heads.
Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}
P(E) = n(E)/n(S)
= 7/8.

An unbiased coin is tossed three times. What is the probability of getting at least two heads?

1. 1/4
2. 5/8
3. 3/4
4. 1/2

Answer (Detailed Solution Below)

Option 4 : 1/2

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Electric charges and coulomb's law (Basic)

10 Questions 10 Marks 10 Mins

Formula Used:

Probability = (Number of successful outcomes/Total number of outcomes)

P(E) = (nE)/(nS), where nE = Number of events and nS = Number of sample space 

Calculation:

When an unbiased coin is tossed three times, we'll get following cases;

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

Total possible outcomes = 8

Here, H = Head, T = Tail

At least 2 heads outcomes = 4

P(E) = (nE)/(nS) = 4/8 = 1/2

∴ Probability of getting at least two heads are 1/2.

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Answer

Verified

Hint : When we toss an unbiased coin we get either a head or tail. Here we are throwing three such coins. We are going to proceed with the thought of getting head and tail sequence on three coins.

Elementary events associated with the random experiment of tossing three coins are HHH, HHT, HTH, THH, HTT, THT, TTH and TTT.
Total number of elementary events = 8
If any of the elementary events HHH, HHT, HTH and THH is an outcome, then we say that the event of “Getting at least two heads” occurs.
$\therefore $ Number of favorable elementary events = 4
$ \Rightarrow P(E) = \frac{{n(E)}}{{n(S)}}$
Hence required probability = $\frac{4}{8} = \frac{1}{2}$

Note:
Probability of an event E is (Number of favorable outcomes to event E) / (Total number of possible outcomes in sample space)
$ \Rightarrow P(E) = \frac{{n(E)}}{{n(S)}}$
In the given problem our event is getting at least two heads while tossing three coins.

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What is the probability of getting 2 heads when 3 unbiased coins are tossed?

p(e)=possible required events/total events. Getting at most Two heads means 0 to 2 but not more than 2. 7/8? P(E)= N(E) /N(S) = 7/8 Ans....

When 3 unbiased coins are tossed once what is the probability of getting all heads?

` <br> `therefore` P (getting all heads) `= P(E_(1)) = (n(E_(1)))/(n(S)) = 1/8.

When a unbiased coin is tossed thrice Where is the probability of getting three heads?

Hence, the probability of getting head all three times = 81.

What is the probability of getting 2 heads when 3 unbiased coins are tossed?

What is the probability of getting at most two heads?

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

When unbiased coin is tossed 3 times what is the probability of getting more than one head?