Math, asked by sravanisai177, 4 months ago

13. The probability of getting at least one head when we toss 3 unbiased coins is​

Answers

Answered by Skyllen
12

Let's assume that H be for head and T for tail.

Total number of outcomes when three unbiased coins are tossed simultaneously:-

  1. HHH,
  2. HHT,
  3. HTH,
  4. HTT,
  5. THH,
  6. THT,
  7. TTH,
  8. TTT

In our given question, we need exact one head, which is appeared in 3 tosses: - HTT, THT and TTH. So these are the possible outcomes for exact one head.

When we tossed 3 coins, there were total 8 equally likely outcomes.

This is observed by taking the number of simultaneous outcomes(2), heads and tails, and then raising it to the power of the number of events(3), 2³ = 2×2×2 = 8

And therefore, probability for exact one head will be 3/8.

Explanation:-

Here, to calculate the probability of getting at least one head or tale, we can have to calculate the probability of getting 1 head, 2 heads and 3 heads and sum them up to get the required probability.

Answered by sara122
4

\huge\tt{\boxed{\underbrace{\overbrace{\bold\color{blue}{❄GiVeN❄}}}}}

  • ㅤㅤㅤㅤㅤ 3 unbiased coins are tossed

 \\  \\  \\

\huge\tt{\boxed{\underbrace{\overbrace{\bold\color{blue}{☛To \:FiNd☚}}}}}

  • Probability of getting one head when 3 coins are toosed simultaneously

 \\  \\  \\

\huge\tt{\boxed{\underbrace{\overbrace{\bold\color{blue}{☛FoRmuLa☚}}}}}

 \large \rm \bold {\boxed {\rm{ \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: p(e) =  \frac{no.of \: trails}{total \: no. \: of \: trails}  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: }}}

 \\  \\  \\

When three unbiased coins are tossed , we get -

ㅤㅤㅤㅤㅤㅤㅤㅤㅤ❄ HHH

HHT

HTH

HTT

TTT

TTH

THT

TTT

 \\  \\  \\

Now , we have to write the one heads when three coins are tossed.

ㅤㅤㅤㅤㅤㅤㅤㅤㅤ⚡THT

HTT

TTH

 \\  \\  \\

By applying the formula:-

 \\ \large\rm \bold{p(e) =  \frac{no. \: of \: one \: head}{total \: no. \: of \:the \: three \: coin \:tossed  } } \\  \\  \\  \large \rm \bold \red{p(of \: getting \: one \: head) =  \frac{3}{8} } \\  \\   \large\rm \bold\therefore{probability \: of \: getting \: one \: head \: when \: three \: coins \: are \: tossed \: is \:  \frac{3}{8} } \\  \\  \\

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