Question

7. A coin is tossed 1000 times with the following frequencies :
Head : 455, Tail : 545
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Answer 1

hey dude

2 coins are tossed so the probablitybof getting heads is

455/1000=91/200

the probability of getting tails is

545/1000=109/200

hop its correct dude

Answer 2

Given :

• A Coin is tossed 1000 times
• Frequency of Head = 455
• Frequency of Tail = 545

To Find :

• The Probability of Each Event.

Formula :

$\sf \red{\boxed{\bf\green{ P(e) \: = \: \dfrac{Number \: of \: trials \: in \: which \: the \: event \: happened}{The \: total \: number \: of \: trials} }}}$

Solution :

✰ Since the coin is tossed 1000 times, the total number of trial is 1000. Let us call the events of getting a head and of a tail as E and F respectively. Then the number of times E happen that is the number of times a head come up is 455.

$\sf { P(e) \: = \: \dfrac{Number \: of \: Heads}{Total \: number \: of \: trials} }$

$\sf { P(e) \: = \: \dfrac{455}{1000} }$

$\bf { P(e) \: = \: 0.455 }\blue \bigstar$

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✰ Similarly, the probability of the event of getting a tail :

$\sf { P(f) \: = \: \dfrac{Number \: of \: Heads}{Total \: number \: of \: trials} }$

$\sf { P(f) \: = \: \dfrac{545}{1000} }$

$\bf { P(f) \: = \: 0.545 }\blue \bigstar$

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Note that in the above question , P(e) + P(f) = 0.455 + 0.545 = 1 , and e and f are the only two possible outcomes of each trial.

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