Dia tosses two different coins simultaneously what is the probability thatshe gets at least one head

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Solution : Let us start with taking Total number of outcomes be=4{HH,TT,TH,HT}<br> Let E=event of getting at least one head<br> Number of favourable outcomes be=3{HT,HH,TH}<br> P(E)=Number of favourable outcomes /Total number of outcomes<br> =`3/4`

Example 9 - Chapter 15 Class 10 Probability (Term 1)

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<p></p><div class="d-flex position-relative justify-content-center align-items-center text-white"><div class="mt-5 black-bg col-md-8 col-sm-12" style="background:#141414"><div class="row m-4 p-4"><div class="pt-1 mb-5"><p class="h2 lh-base text-center" style="color:#fff">Maths Crash Course - Live lectures + all videos + Real time Doubt solving!</p></div><div class="col-12 justify-content-center text-center mt-3"><a target="_blank" href="//www.teachoo.com/premium/maths/crash-course/" class="btn join-now-button fs-5 stretched-link">Join Maths Crash Course now</a></div></div></div></div> </div> </div> <div class="card p-3 m-3 align-items-center text-center"> <hr> <h3>Transcript</h3><span class="ezoic-autoinsert-video ezoic-under_first_paragraph"></span> <p id="transcript" itemprop="articleBody">Example 9 Harpreet tosses two different coins simultaneously (say, one is of Re 1 and other of Rs 2). What is the probability that she gets at least one head? Total possible outcomes = (H, H) , (H, T), (T, H), (T, T) Number of possible outcomes = 4 Total outcomes where she get atleast one head = (H, H) , (H, T), (T, H) Number of outcomes where she get atleast one head = 3 P(she gets atleast one head) = (π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘œπ‘’π‘‘π‘π‘œπ‘šπ‘’π‘  π‘€β„Žπ‘’π‘Ÿπ‘’ π‘ β„Žπ‘’ 𝑔𝑒𝑑𝑠 π‘Žπ‘‘π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’ β„Žπ‘’π‘Žπ‘‘)/(π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘π‘œπ‘ π‘ π‘–π‘π‘™π‘’ π‘œπ‘’π‘‘π‘π‘œπ‘šπ‘’π‘ ) = 3/4</p> <button class="btn btn-outline-dark" id="show-more">Show More</button> </div> <div class="bottom-post-buttons d-flex"> <a target="_blank" id="next-button" title="ga('send','event','Next button on post','Click','Example 9 - Chapter 15 Class 10 Probability (Term 1)');" class="btn btn-success next" href="//www.teachoo.com/8963/564/Example-10--(Optional)/category/Examples/"><b>Next</b>: Example 10* (Optional) β†’ </a> <a type="button" class="btn btn-info ms-4 px-2 py-3" data-bs-toggle="modal" data-bs-target="#blackQuesModal" title="This is a Teachoo Black feature"> Ask a doubt (live) <svg xmlns="//www.w3.org/2000/svg" width="16" height="16" fill="currentColor" class="bi bi-magic" viewbox="0 0 16 16"> <path d="M9.5 2.672a.5.5 0 1 0 1 0V.843a.5.5 0 0 0-1 0v1.829Zm4.5.035A.5.5 0 0 0 13.293 2L12 3.293a.5.5 0 1 0 .707.707L14 2.707ZM7.293 4A.5.5 0 1 0 8 3.293L6.707 2A.5.5 0 0 0 6 2.707L7.293 4Zm-.621 2.5a.5.5 0 1 0 0-1H4.843a.5.5 0 1 0 0 1h2.829Zm8.485 0a.5.5 0 1 0 0-1h-1.829a.5.5 0 0 0 0 1h2.829ZM13.293 10A.5.5 0 1 0 14 9.293L12.707 8a.5.5 0 1 0-.707.707L13.293 10ZM9.5 11.157a.5.5 0 0 0 1 0V9.328a.5.5 0 0 0-1 0v1.829Zm1.854-5.097a.5.5 0 0 0 0-.706l-.708-.708a.5.5 0 0 0-.707 0L8.646 5.94a.5.5 0 0 0 0 .707l.708.708a.5.5 0 0 0 .707 0l1.293-1.293Zm-3 3a.5.5 0 0 0 0-.706l-.708-.708a.5.5 0 0 0-.707 0L.646 13.94a.5.5 0 0 0 0 .707l.708.708a.5.5 0 0 0 .707 0L8.354 9.06Z"> </path> </svg> </a> </div> <nav aria-label="breadcrumb"> <ol class="bottom-breadcrumb" id="bottom-breadcrumb"> <li><a target="_blank" class="breadcrumb-value" href="//www.teachoo.com/subjects/cbse-maths/class-10th/ch25-10th-probability/">Chapter 15 Class 10 Probability</a></li> <li>Serial order wise</li> <ins class="adsbygoogle" style="display:block" data-ad-client="ca-pub-9822854285786923" data-ad-slot="2438274818" data-ad-format="auto" data-full-width-responsive="true"></ins><script>(adsbygoogle = window.adsbygoogle || []).push({});

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