Total number of cases = 52 Total face cards = 12 [favourable cases] So **probability** = 12 /52 = 3/13 3. There is a pack of 52 cards and Rohan draws **two** cards **together**, what is the **probability** that **one** is spade and **one** is heart ?.

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Given, a **coin** is **tossed** 3 times. We have to **find** **the** **probability** **of** **getting** **at** **least** **one** **head**. **The** possible outcomes **are**. T H H. H T H. H H T. T T H. T H T. H T T. T T T. H H H. We observe that there is only **one** scenario in throwing all **coins** where there are no **heads**. **The** chances for **one** given **coin** to be **heads** is 1/2.

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Solution: We know that, when **two** **coins** **are** **tossed** **together** possible number of outcomes = {HH, TH, HT, TT} So, \ [n\left ( S \right)\text { }=\text { }4\] \ [\left ( i \right)\]E = event of **getting** both **heads** = {HH} \ [n\left ( E \right)\text { }=\text { }1\].

First total possibilities 8 = 2 x 2 x 2 Second **Probability** **of** **Head** 50% (0.5) so 3 **coin** flips 1.5 = 0.5 + 0.5 + 0.5 That gives you the **probability** **of** 1 **head** so double it for 2 **heads** is 3 = 1.5 x 2 (**Heads**) So 0.375 = 3/8 or 37.5% I hope that helps anyone who doesn't want to write out all the possibilities by hand. $\endgroup$ -.

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If three **coins** **are** **tossed** simultaneously, then write their outcomes.$( a)$. All possible outcomes.$( b)$. Number of possible outcomes.$( c)$. **Find** **the** **probability** **of** **getting** **at** **least** **one** **head**.$( d)$. **Find** **the** **Probability** **of** **getting** **at** most **two** **heads**.$( e)$. **Find** **the** **Probability** **of** **getting** no tails.

Three **coins** **are** **tossed** **together**. **Find** **the** **probability** **of** **getting**: (i) exactly **two** **heads** (ii) at most **two** **heads** (iii) at **least** **one** **head** and **one** tail asked Mar 19, 2020 in **Probability** by ShasiRaj ( 62.5k points). Question **Two** **coins** **are** **tossed** simultaneously. **Find** **the** **probability** **of** **getting** i) exactly **one** **head** ii) at **least** **one** tail iii) no tail iv) at most **one** **head** Medium Solution Verified by Toppr When **two** **coins** **are** **tossed** **the** results are (HH,HT,TH,TT) Total no. of outcomes =4 1) Exactly **one** **head** = only **two** cases are (HT,TH) =42=21.

step 1 **Find** **the** total possible combinations of sample space S S = {HH, HT, TH, TT} S = 4 step 2 **Find** **the** expected or successful events A A = {HT, TH} A = 2 step 3 **Find** **the** **probability** P (A) = Successful Events Total Events of Sample Space = 2 4 = 0.5 P (A) = 0.5 0.5 is the **probability** **of** **getting** exactly 1 **Head** in 2 tosses.

Total number of cases = 52 Total face cards = 12 [favourable cases] So **probability** = 12 /52 = 3/13 3. There is a pack of 52 cards and Rohan draws **two** cards **together**, what is the **probability** that **one** is spade and **one** is heart ?.

Answer (**1** of 5): This is an example of a binomial distribution: k = number of trials = 7 n = number of successes (**heads**) = **2** p = **probability** of success = .5 q = **probability** of failure = .5 P(x = n) = C(k, n)p^n q^{k-n} P(x = **2**) = C(7, **2**)(0.5)^**2** (0.5)^{7–**2**} P(x = **2**) = 21 \cdot.

**Two coins are tossed once. Find the probability** ofi **Getting 2** headsii **Getting at least 1** headiii **Getting** no headiv **Getting 1 head** and **1** tail Uh-Oh! That’s all you get for now. We would love to personalise your learning journey. Sign Up to explore more. Sign Up or.

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**Two coins are tossed once. Find the probability** ofi **Getting 2** headsii **Getting at least 1** headiii **Getting** no headiv **Getting 1 head** and **1** tail Uh-Oh! That’s all you get for now. We would love to personalise your learning journey. Sign Up to explore more. Sign Up or.

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Three **coins** **are** **tossed** **together**. **Find** **the** **probability** **of** **getting** : (i) exactly **two** **heads** (ii) at **least** **two** **heads** (iii) at **least** **one** **head** and **one** tail (iv) no tails Solution Total number of events **tossed** by 3 **coins** each having **one** **head** and **one** tail = 2×2×2 =8.

Complete step-by-step answer: If **two** unbiased **coins** **are** **tossed** simultaneously, then the total number of possible outcomes may be either. 1.both **head** HH. 2.one **head** and **one** tail (HT, TH) 3.Both tail (TT) So here total number of possible cases = 4. (i) **two** **heads**. a favourable outcome for **two** **heads** is HH.

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question_answer Answers (2) person Raghunath Reddy When 3 unbaised **coins** **are** **tossed**, **the** total number of events = 8 They are {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT} (1) Number of favourable events (**getting** **at** **least** **two** **heads**) = 4 **Probability** **of** **getting** **at** **least** **two** **heads** = 4/8 = 1/2 (2) At most **two** **heads** That is no **head** or **one** **head** or 2 **heads**.

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Solution ∵ A **coin** has **two** faces **Head** and Tail or H, T ∴ **Two** **coins** **are** **tossed** ∴ Number of **coins** = 2 x 2 = 4 which are HH, HT, TH, TT (i) At **least** **one** **head**, then Number of outcomes = 3 ∴ P (E) = Number of favourable outcome Number of all possible outcome Number of favourable outcome Number of all possible outcome = 3 4 (ii).

Question Three unbiased **coins** **are** **tossed** **together**. **Find** **the** **probability** **of** **getting** (a) (i) **one** tail(ii) **two** tails(iii) all tails(iv) at **least** **two** tails (b) (i) at most **two** tails(ii) At most **two** **heads** Medium Solution Verified by Toppr Here S= ( TTT,TTH,THT,HTT,THH,HTH,HHT,HHH ) a (i) **one** tail (THH,HTH,HHH) Then E=3 So (p)= n(S)n(E)= 83.

VIDEO ANSWER:four fair **coins** **are** **tossed** **together**. For fear **coins** toast **together**. What's the **probability** **of** **getting** **at** **least** three **heads** **probability** **of** **getting** **at** **least** three **head**. So the total number of spaces total sample spaces That will be through the power of four. That is 16. We have to **find** **the** **probability** that is at **least** **Getting** **at**.

TO **FIND**: **Probability of getting at least one head**. When **two coins are tossed** then the outcome will be. TT, HT, TH, HH. Hence total number of outcome is 4. **At least one head** means 1H or 2H. Hence total number of favorable outcome i.e. **at least one head** is.

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Three unbiased **coins** **are** **tossed** **together**. **find** **the** **probability** **of** **getting** (i) **one** **head** (ii) **two** **heads** (iii) all **heads** (iv) at **least** **two** **heads**. - 2889633.

Three fair **coins are tossed together**. **Find the probability of getting** (i) all **heads** (ii) **atleast one** tail (iii) atmost **one head** asked Oct 23, 2020 in Statistics and **Probability** by Darshee ( 49.1k points).

If **Two** **Coins** **Are** **Tossed**, **Find** **the** **Probability** **of** **the** Following Event.**Getting** **at** **Least** **One** **Head**. Maharashtra State Board SSC (English Medium) 10th Standard Board Exam. Question Papers 257. Textbook Solutions 13091. MCQ Online Tests 39. Important Solutions 3201.

Answer (1 of 22): When **two** **coins** **are** flipped the possible outcomes are (H,H),(H,T),(T,H)(T,T) Favorable out come is (TT) Probability=1/4. Information about this question covers all topics & solutions for Class 9 2022 Exam. **Find** important definitions, questions, meanings, examples, exercises and tests below for **Two** **coins** **are** **tossed** simultaneously. The **probability** **of** **getting** **at** **least** **one** **head** isa)1/2b)1/4c)0d)3/4Correct answer is option 'D'. Can you explain this answer?.

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Users may refer the below solved example work with steps to learn how to **find** what is **the probability of getting at-least 2 heads**, if a **coin** is **tossed** fix times or 6 **coins tossed together**. Users may refer this tree diagram to learn how to **find** all the possible combinations of sample space for flipping a **coin one**, **two**, three or four times.

Thus, the **probability** **of** **getting** **at** **least** **two** **heads** when three **coins** **are** **tossed** simultaneously = 4/8 = 1/2 (iii) For **getting** **at** **least** **one** **head** and **one** tail the cases are THT, TTH, THH, HTT, HHT, and HTH. So, the total number of favourable outcomes i.e. at **least** **one** tail and **one** **head** is 6.

**Probability** **of** **getting** exactly **two** **heads** = 3/8 (b)at most **two** **heads** Favourable conditions =TTT,TTH,THT,THH,HTT HTH,HHT = 7 **Probability** **of** **getting** **at** most **two** **heads** = 7/8 C)at **least** **one** **head** and **one** tail: Favourable conditions:TTH,THT,THH,HTT HTH,HHT = 6 **Probability** **of** **getting** **at** **least** **one** **head** and **one** tail = 6/8 = 3/4 d)no tails:. Users may refer the below solved example work with steps to learn how to **find** what is **the probability of getting at-least 2 heads**, if a **coin** is **tossed** four times or 4 **coins tossed together**. Users may refer this tree diagram to learn how to **find** all the possible combinations of sample space for flipping a **coin one**, **two**, three or four times.

**The** outcomes of the event, **getting** **at** **least** **two** **heads** = HHT, HTH , THH, HHH. ∴Number of outcomes of the event **getting** **at** **least** **two** **heads** by tossing three unbiased **coins** = 4. Thus, the **probability** **of** **getting** **at** **least** **two** **heads**= **The** outcomes of the event **getting** **at** most **two** **heads** = HHT, HTH , HTT,THH,THT,TTH,TTT.

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Answer (1 of 47): Back to basics - we have **two** independent events. To get the joint **probability** (i.e. "AND"), **one** multiplies the individual probabilities. So first **one** has to figure out the **probability** **of** **getting** a **head** when tossing **one** **coin**. (That can be done by understanding that the probabilit.

Answer (1 of 22): When **two** **coins** **are** flipped the possible outcomes are (H,H),(H,T),(T,H)(T,T) Favorable out come is (TT) Probability=1/4.

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**The** ratio of successful events A = 11 to the total number of possible combinations of a sample space S = 16 is the **probability** **of** 2 **heads** in 4 **coin** tosses. Users may refer the below solved example work with steps to learn how to **find** what is the **probability** **of** **getting** **at-least** 2 **heads**, if a **coin** is **tossed** four times or 4 **coins** **tossed** **together**.

Answer (**1** of 4): There are 4 possible outcomes (Sample space)— HH,HT,TH,TT. If you want both **heads** then the favourable outcome is HH and **the probability** is **1**/4. If you want **atleast one head** then **the probability** can be calculated as follows— P(**atleast one head**.

What is the **probability** **of** **getting** **at** **least** **one** **head** and **one** tail? This question was previously asked in. ... When three **coins** **are** **tossed**, total possible outcomes = 8. S = {HHH, HHT, HTT, THH, TTH, THT, HTH} ... Given below are **two** statements : Statement I : If **two** trains of length x km and y km are moving in the same direction at u kmph and v.

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**The** ratio of successful events A = 31 to the total number of possible combinations of a sample space S = 32 is the **probability** **of** 1 **head** in 5 **coin** tosses. Users may refer the below solved example work with steps to learn how to **find** what is the **probability** **of** **getting** **at-least** 1 **head**, if a **coin** is **tossed** five times or 5 **coins** **tossed** **together**.

**Find** an answer to your question 3 **coins are tossed together** . **find the probability of getting** : **2 heads**. **at least 2 heads**. at most **2 heads**. at most **1** tail. at susanna1 susanna1 03.09.2018 Math Secondary School answered • expert verified.

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**Two coins are tossed Find: the probability that** only **one head** is obtained. Solution: When **two coins are tossed** then all possible outcomes are HH , HT , TH and TT. Total number of outcomes = 4 • • P(**getting one head**) = **2**/4 Therefore, **the probability** that only #.

**Two coins are tossed together**. **The probability of getting** exactly **one head** is (a) 1/4 (b) 1/2 (c) 3/4 (d) 1.

Three unbiased **coins** **are** **tossed** **together**. **Find** **the** **probability** **of** **getting**: (i) all **heads**. (ii) exactly **two** **head**. (iii) exactly **one** **heads**. (iv) at **least** **two** **heads**. (v) at **least** **two** tails. **probability** class-10 1 Answer 0 votes answered Feb 9, 2018 by Vikash Kumar (257k points) Best answer.

Solution ∵ A **coins** has **two** faces **Head** and Tail or H, T ∴ **Two** **coins** **are** toosed ∴ Number of **coins** = 2×2 =4 which are HH, HT, TH, TT (i) At **least** **one** **head**, then Number of outcomes= 3 ∴ P (E) = Number of favourable outcome Number of all possible outcome = 3 4 (ii) When both **head** or both tails, then Number of outcomes=2.

Example : Suppose we have 3 unbiased **coins** and we have to **find** **the** **probability** **of** **getting** **at** **least** 2 **heads**, so there are 2 3 = 8 ways to toss these **coins**, i.e., HHH, HHT, HTH, HTT, THH, THT, TTH, TTT Out of which there are 4 set which contain at **least** 2 **Heads** i.e., HHH, HHT, HH, THH So the **probability** is 4/8 or 0.5.

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**The** **probability** **of** **getting** 2 **heads** with 2 **coins** can be broken into 2 steps: The first **coin** must show **Heads** AND the second **coin** must show **Heads**. P (H,H) = P (H) AND P (H) ← 'and' means multiply P (H) × P (H) = 1 2 × 1 2 = 1 4 Answer link.

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Total number of outcomes possible when a **coin** is **tossed** = 2 (? **Head** or Tail)Hence, total number of outcomes possible when 5 **coins** **are** **tossed**, n(S) = 25E = Event of **getting** exactly 2 **heads** when 5 **coins** **are** tossedn(E) = Number of ways of **getting** exactly 2 **heads** when 5 **coins** **are** **tossed** =5C2.

I **know** if the question said: **at least one head** then I would do: ${5\choose0}=**1**$ $**2**^5=32-**1** = 31 $ Stack Exchange Network Stack Exchange network consists of 182 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

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When **two** **of** **the** unbiased **coins** **are** **tossed**, what is the **probability** **of** both showing **head**? Solution: As given in the question, No of unbiased **coins** **are** **tossed** = 2 The **probability** **of** showing both **heads** is P (E) = No of favourable outcomes/Total no of outcomes Total no of outcomes is represented with sample space (S) S = {HH, HT, TH, TT} n (S) = 4.

If we toss **two coins** simultaneously, then possible out comes (s), are S = { HT, TH, HH, TT } ⇒ n( S) = 4 Let E be the favourable outcomes **of getting two heads**, then E = { H H } ⇒ n(E) = **1** Therefore, P(E) = Let F be the favourable outcomes **of getting at least**.

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**Two coins are tossed Find: the probability that** only **one head** is obtained. Solution: When **two coins are tossed** then all possible outcomes are HH , HT , TH and TT. Total number of outcomes = 4 • • P(**getting one head**) = **2**/4 Therefore, **the probability** that only #.

Example : Suppose we have 3 unbiased **coins** and we have to **find** **the** **probability** **of** **getting** **at** **least** 2 **heads**, so there are 2 3 = 8 ways to toss these **coins**, i.e., HHH, HHT, HTH, HTT, THH, THT, TTH, TTT Out of which there are 4 set which contain at **least** 2 **Heads** i.e., HHH, HHT, HH, THH So the **probability** is 4/8 or 0.5.

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According to the question, three **unbiased coins are tossed simultaneously**. Now, by unbiased **coins**, we mean that those **coins** which have equal **probability of getting** either a **head** or a tail. Now, we have to **find the probability** of various events. n ( S) shows the total number of outcomes, n ( S) = 8, as this is the sample space.

Total number of cases = 52 Total face cards = 12 [favourable cases] So **probability** = 12 /52 = 3/13 3. There is a pack of 52 cards and Rohan draws **two** cards **together**, what is the **probability** that **one** is spade and **one** is heart ?.

An Unbiased **coin** is **tossed** 4 times. What is the **Probability** **of** **getting** (i) 3 **heads** Q. If (a - b + c) : (b - c + 2d) : (2a + c - d) = 2 : 3 : 5, then **find** (3a + 3c - 2d) : d. Q. write a letter to your sister for taking part in a such science exhibition to have an experience for better future as its an era of science encourage her for.

Best answer Three fair **coins** **are** **tossed** **together** Sample spade = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} n (S) = 8 (i) Let A be the event of **getting** all **heads** A = {HHH} n (A) = 1 P (A) = n(A) n(S) = 1 8 n ( A) n ( S) = 1 8 (ii) Let B be the event of **getting** **atleast** **one** tail. B = {HHT, HTH, HTT, THH, THT, TTH, TTT} n (B) = 7.

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Users may refer the below solved example work with steps to learn how to **find** what is the **probability** **of** **getting** **at-least** 2 tails, if a **coin** is **tossed** three times or 3 **coins** **tossed** **together**. Users may refer this tree diagram to learn how to **find** all the possible combinations of sample space for flipping a **coin** **one**, **two**, three or four times.

Suppose 5 fair **coins** **are** **tossed** simultaneously. What is the probablity of **getting** a various number of **heads**? ... **Probability** **of** **Head** in **coin** flip when **coin** is flipped **two** times. 0. **Probability** **of** tossing five **coins** and **getting** **at** **least** **one** **head**. 0. ... Three people each flip **two** fair **coins**.**Find** **the** **probability** that exactly **two** **of** **the** people.

**Find the probability of getting** more **heads** than the number of tails. asked Dec 22, 2020 in **Probability** by Gaangi (24.9k points) **probability** class-9 0 votes **1** answer Three unbiased **coins are tossed** once. **Find the probability of getting** at most **2** tails or **at least 2** ).

With 5 **coins** to flip you just times 16 by 2 and then minus 1, so it would result with a 31 in 32 chance of **getting** **at** **least** **one** **heads**. With 6 **coins** you times by 2 and minus by 1 again resulting in a 63 in 64 chance. To **find** **the** chance of **getting** **at** **least** **one** **heads** if you flip ten **coins** you times 64 by 2 four times or by 16 once and then minus 1.

"Hey man, but girls and **coins** **are** **two** different things! I should know, I've seen at **least** **one** **of** each." Well, let me explain that these **two** problems are basically the same, that is, from the point of view of mathematics.Whether you want to toss a **coin** or ask a girl out, there are only **two** possibilities that can occur.In other words, if you assign the success of your experiment, be it **getting**.

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On tossing a **coin** twice, the possible outcomes are {HH, TT, HT, TH} Therefore, the total number of outcomes is 4 **Getting** only **one** tail includes {HT, TH} Therefore, the number of favorable outcomes is 2 Hence, the **probability** **of** **getting** exactly **one** tail is 2/4 = 1/2 1/2 Important Notes.

Answer (1 of 47): Back to basics - we have **two** independent events. To get the joint **probability** (i.e. "AND"), **one** multiplies the individual probabilities. So first **one** has to figure out the **probability** **of** **getting** a **head** when tossing **one** **coin**. (That can be done by understanding that the probabilit.

If we toss **two** **coins** simultaneously, then possible out comes (s), **areS** = { HT, TH, HH, TT }⇒ n( S) = 4Let E be the favourable outcomes of **getting** **two** **heads**, thenE = { H H }⇒ n(E) = 1Therefore, P(E) = Let F be the favourable outcomes of **getting** **at** **least** **one** **head**, thenF = { HH, HT, TH }⇒ n(F) = 3Therefore, P(F) = Let G be the favourable outcomes of **getting** no **head** then⇒ n (G.

Answer (**1** of 47): Back to basics - we have **two** independent events. To get the joint **probability** (i.e. “AND”), **one** multiplies the individual probabilities. So first **one** has to figure out **the probability of getting** a **head** when tossing **one coin**. (That can be done by.

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Three **coins** **are** **tossed** **together**. **Find** **the** **probability** **of** **getting**: (i) exactly **two** **heads** (ii) at most **two** **heads** (iii) at **least** **one** **head** and **one** tail asked Mar 19, 2020 in **Probability** by ShasiRaj ( 62.5k points). Best answer Three fair **coins** **are** **tossed** **together** Sample spade = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} n (S) = 8 (i) Let A be the event of **getting** all **heads** A = {HHH} n (A) = 1 P (A) = n(A) n(S) = 1 8 n ( A) n ( S) = 1 8 (ii) Let B be the event of **getting** **atleast** **one** tail. B = {HHT, HTH, HTT, THH, THT, TTH, TTT} n (B) = 7.

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Three fair **coins are tossed together**. **Find the probability of getting** (i) all **heads** (ii) **atleast one** tail (iii) atmost **one head** asked Oct 23, 2020 in Statistics and **Probability** by Darshee ( 49.1k points).

A **coin** is **tossed** **two** times. **Find** **the** **probability** **of** **getting** not more than **one** **head**. (2011D) Solution: S = {HH, HT, TH, TT} = 4 Favourable cases are HT, TH, TT i.e., 3 cases ∴ P (not more than 1 **head**) = \(\frac{3}{4}\) Question 23. Three distinct **coins** **are** **tossed** **together**. **Find** **the** **probability** **of** **getting** (2015 D) (i) at **least** 2 **heads** (ii) at.

Three fair **coins** **are** **tossed**. What is the **probability** **of** **getting** three **heads** given that at **least** **two** **coins** show **heads**? Maharashtra State Board HSC Arts 11th. Textbook Solutions 9042. Important Solutions 4. Question Bank Solutions 5433. Concept Notes & Videos 409. Syllabus.

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Best answer Three fair **coins** **are** **tossed** **together** Sample spade = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} n (S) = 8 (i) Let A be the event of **getting** all **heads** A = {HHH} n (A) = 1 P (A) = n(A) n(S) = 1 8 n ( A) n ( S) = 1 8 (ii) Let B be the event of **getting** **atleast** **one** tail. B = {HHT, HTH, HTT, THH, THT, TTH, TTT} n (B) = 7.

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So the cases are **head**, **head** **head** Had the tail had the had the tail tail Still still had the still had the pill Still had had the had had a pill So there are seven cases for at **least** **one** **head**. So **probability** off **getting** **at** **least** **one** had is comes out Toby seven upon it. Now we need to **find** labrum lt forgetting exactly **two** tails.

When **the** **coin** is **tossed** 2 times, the possible outcomes are {TH, HT, TT, HH}. Since in 3 out of 4 outcomes, **heads** don't occur **together**. Therefore, the required **probability** is (3/4) or 0.75. Input: N = 3 Output: 0.62 Explanation: When the **coin** is **tossed** 3 times, the possible outcomes are {TTT, HTT, THT, TTH, HHT, HTH, THH, HHH}.

A **coin** is **tossed** **two** times. **Find** **the** **probability** **of** **getting** **at** **least** **one** **head**. Solution: Total outcomes are 4, i.e. (HH, HT, TH, TT} Favourable outcomes are 3, i.e. {HH, HT and TH} P(at **least** **one** head)=3/4. Question 67. A **coin** is **tossed** **two** times. **Find** **the** **probability** **of** **getting** both **heads** or both tails. Solution:.

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