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Plz ans this question

Q.4. Three coins are tossed 6 times and outcomes are observed as (HHT), (HTT), (THT), (TTT), (THH), (HHH). Prepare a frequency table on the basis of numbers of heads and hence calculate the percentage chance of occurrence.

Please find below the solution to the asked query :

$\mathrm{Percentage}\mathrm{chance}\mathrm{of}\mathrm{occurance}\mathrm{of}0\mathrm{head}=\frac{1}{6}\times 100=16.66\%\phantom{\rule{0ex}{0ex}}\mathrm{Percentage}\mathrm{chance}\mathrm{of}\mathrm{occurance}\mathrm{of}1\mathrm{head}=\frac{2}{6}\times 100=33.33\%\phantom{\rule{0ex}{0ex}}\mathrm{Percentage}\mathrm{chance}\mathrm{of}\mathrm{occurance}\mathrm{of}2\mathrm{heads}=\frac{2}{6}\times 100=33.33\%\phantom{\rule{0ex}{0ex}}\mathrm{Percentage}\mathrm{chance}\mathrm{of}\mathrm{occurance}\mathrm{of}3\mathrm{heads}=\frac{1}{6}\times 100=16.66\%$

Number of heads | Frequency |

0 | 1 |

1 | 2 |

2 | 2 |

3 | 1 |

$\mathrm{Percentage}\mathrm{chance}\mathrm{of}\mathrm{occurance}\mathrm{of}0\mathrm{head}=\frac{1}{6}\times 100=16.66\%\phantom{\rule{0ex}{0ex}}\mathrm{Percentage}\mathrm{chance}\mathrm{of}\mathrm{occurance}\mathrm{of}1\mathrm{head}=\frac{2}{6}\times 100=33.33\%\phantom{\rule{0ex}{0ex}}\mathrm{Percentage}\mathrm{chance}\mathrm{of}\mathrm{occurance}\mathrm{of}2\mathrm{heads}=\frac{2}{6}\times 100=33.33\%\phantom{\rule{0ex}{0ex}}\mathrm{Percentage}\mathrm{chance}\mathrm{of}\mathrm{occurance}\mathrm{of}3\mathrm{heads}=\frac{1}{6}\times 100=16.66\%$

Regards

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